Question

A rectangular tank of height 10 m filled with water, is placed near the bottom of an incline of angle 30°. At height x from the bottom, a small hole is made as shown in figure such that the stream coming out from the hole strikes the inclined plane normally. Find the value of 0.6 × ?

Answer 1

5

Answer 2

The velocity of the water stream coming out of the hole at height x from the bottom is given by Torricelli's theorem:

$v_0 = \sqrt{2g(10-x)}$

The hole is at height x from the bottom of the tank. The height of water above the hole is $10-x$.

Let's set up a coordinate system with the origin at the hole. The x-axis is horizontal and points away from the tank, and the y-axis is vertical downwards.

The initial velocity of the stream is purely horizontal: $u_x = v_0 = \sqrt{2g(10-x)}$, $u_y = 0$.

The acceleration due to gravity is $g$ in the positive y-direction: $a_x = 0$, $a_y = g$.

The position of a point on the stream at time t is given by:

$X(t) = u_x t + \frac{1}{2} a_x t^2 = \sqrt{2g(10-x)} t$

$Y(t) = u_y t + \frac{1}{2} a_y t^2 = 0 \cdot t + \frac{1}{2} g t^2 = \frac{1}{2} g t^2$

The velocity of the stream at time t is given by:

$v_x(t) = u_x + a_x t = \sqrt{2g(10-x)}$

$v_y(t) = u_y + a_y t = 0 + gt = gt$

The velocity vector is $\vec{v}(t) = \sqrt{2g(10-x)} \hat{i} + gt \hat{j}$.

The inclined plane makes an angle of 30° with the horizontal. The stream strikes the inclined plane normally. This means the velocity vector of the stream at the point of impact is perpendicular to the inclined plane.

The slope of the inclined plane is $\tan(30^\circ)$. The angle of the inclined plane with the horizontal is 30°.

A line normal to the inclined plane makes an angle of $90^\circ + 30^\circ = 120^\circ$ or $30^\circ - 90^\circ = -60^\circ$ with the horizontal.

Since the stream is falling downwards and moving horizontally away from the tank, the velocity vector will be in the first quadrant of the (X, Y) coordinate system (with Y downwards). The angle the velocity vector makes with the positive X-axis is $\alpha$, where $\tan \alpha = \frac{v_y}{v_x} = \frac{gt}{\sqrt{2g(10-x)}}$. This angle is positive.

The velocity vector is perpendicular to the inclined plane. The inclined plane is below the hole. The normal vector to the inclined plane pointing towards the stream's path will be in the direction that makes an angle of $90^\circ - 30^\circ = 60^\circ$ with the vertical downwards direction, or $30^\circ$ with the horizontal. This is incorrect.

Let's consider the angle the velocity vector makes with the horizontal. Let this angle be $\theta$.

$v_x = \sqrt{2g(10-x)}$

$v_y = gt$

The velocity vector is $\vec{v} = v_x \hat{i} + v_y \hat{j}$. The angle $\theta$ is given by $\tan \theta = \frac{v_y}{v_x} = \frac{gt}{\sqrt{2g(10-x)}}$. This angle is measured below the horizontal.

The inclined plane makes an angle of 30° with the horizontal.

The stream strikes the inclined plane normally. This means the velocity vector is perpendicular to the plane.

The angle between the velocity vector and the horizontal is $\theta$. The angle between the inclined plane and the horizontal is 30°.

The condition for the velocity vector to be normal to the inclined plane is that the angle between the velocity vector and the horizontal plus the angle between the inclined plane and the horizontal is 90°.

From the geometry, the angle between the velocity vector (which is below the horizontal) and the inclined plane (which is above the horizontal) being 90 degrees means that the angle the velocity vector makes with the horizontal must be $90^\circ - 30^\circ = 60^\circ$.

So, $\theta = 60^\circ$.

$\tan \theta = \tan 60^\circ = \sqrt{3}$.

Therefore, $\frac{gt}{\sqrt{2g(10-x)}} = \sqrt{3}$.

Squaring both sides: $\frac{g^2 t^2}{2g(10-x)} = 3$.

$\frac{g t^2}{2(10-x)} = 3$.

$g t^2 = 6(10-x)$.

Now consider the position of the striking point. Let the inclined plane pass through the point $(X_0, Y_0)$ relative to the hole. The equation of the inclined plane is $Y - Y_0 = \tan(30^\circ) (X - X_0)$. However, we don't know $(X_0, Y_0)$.

Let's consider the coordinates relative to the bottom of the tank. Let the bottom of the tank be at $(0, 0)$. The hole is at $(0, x)$. The inclined plane starts at some point on the x-axis, say $(d, 0)$, and goes upwards. The equation of the inclined plane is $y = \tan(30^\circ) (z - d)$ for $z \ge d$. This doesn't seem to fit the figure.

Let's go back to the coordinate system with the origin at the hole, x-axis horizontal, y-axis vertical downwards.

Position: $X(t) = \sqrt{2g(10-x)} t$, $Y(t) = \frac{1}{2} g t^2$.

Velocity: $v_x = \sqrt{2g(10-x)}$, $v_y = gt$.

The slope of the velocity vector is $\frac{v_y}{v_x} = \frac{gt}{\sqrt{2g(10-x)}}$.

The inclined plane makes an angle of 30° with the horizontal. The slope of the inclined plane is $\tan(30^\circ) = \frac{1}{\sqrt{3}}$.

The stream strikes the plane normally, so the product of the slope of the velocity vector and the slope of the inclined plane is -1 (if both are considered in the same coordinate system where positive y is upwards).

Let's use the coordinate system where x is horizontal and y is vertical upwards, with origin at the hole.

Initial velocity: $u_x = \sqrt{2g(10-x)}$, $u_y = 0$.

Velocity at time t: $v_x = \sqrt{2g(10-x)}$, $v_y = -gt$.

The slope of the velocity vector is $m_v = \frac{v_y}{v_x} = \frac{-gt}{\sqrt{2g(10-x)}}$.

The inclined plane makes an angle of 30° with the horizontal. The slope of the inclined plane is $m_p = \tan(30^\circ) = \frac{1}{\sqrt{3}}$.

The condition for the velocity vector to be normal to the inclined plane is $m_v m_p = -1$.

$(\frac{-gt}{\sqrt{2g(10-x)}})(\frac{1}{\sqrt{3}}) = -1$.

$\frac{gt}{\sqrt{3}\sqrt{2g(10-x)}} = 1$.

$gt = \sqrt{3}\sqrt{2g(10-x)}$.

Squaring both sides: $g^2 t^2 = 3 \cdot 2g(10-x)$.

$gt^2 = 6(10-x)$.

Now we need to relate the position $(X, Y)$ of impact to the equation of the inclined plane.

Let's assume the inclined plane starts at the horizontal level of the bottom of the tank. The hole is at height x from the bottom. So, the vertical position of the hole is x relative to the bottom.

Let's use a coordinate system with the origin at the bottom of the tank, x-axis horizontal, y-axis vertical upwards.

The hole is at $(0, x)$. Initial velocity is $\vec{u} = \sqrt{2g(10-x)} \hat{i}$.

Position at time t: $\vec{r}(t) = (\sqrt{2g(10-x)} t) \hat{i} + (x - \frac{1}{2}gt^2) \hat{j}$.

Let the inclined plane pass through the origin $(0, 0)$. The equation of the inclined plane is $y = x \tan(30^\circ)$. This doesn't make sense as the trajectory is below the hole level initially.

Let's assume the inclined plane starts at some horizontal distance d from the tank at the bottom level. The equation of the inclined plane is $y = (z-d) \tan(30^\circ)$ for $z \ge d$.

The position of the stream is $(X(t), Y(t))$ relative to the bottom of the tank.

$X(t) = \sqrt{2g(10-x)} t$

$Y(t) = x - \frac{1}{2}gt^2$

The stream hits the plane when $Y(t) = (X(t)-d) \tan(30^\circ)$.

$x - \frac{1}{2}gt^2 = (\sqrt{2g(10-x)} t - d) \tan(30^\circ)$.

Let's use the geometry from the similar question's solution.

In the similar question, the coordinate system is taken with the origin at the hole, x-axis horizontal, y-axis vertically upwards.

Velocity at time t: $v_x = \sqrt{2g(10-h)}$, $v_y = -gt$.

The angle the velocity vector makes with the horizontal is $\phi$, $\tan \phi = \frac{v_y}{v_x} = \frac{-gt}{\sqrt{2g(10-h)}}$.

The inclined plane makes an angle of 30° with the horizontal.

The stream strikes the plane normally. The velocity vector is perpendicular to the plane.

The angle between the velocity vector and the horizontal is $\phi$. The angle between the plane and the horizontal is 30°.

From the figure, the velocity vector is downwards and to the right. The inclined plane is upwards and to the right.

The angle of the velocity vector with the horizontal is $|\phi|$. The angle of the inclined plane with the horizontal is 30°.

The angle between the velocity vector and the inclined plane is 90°.

The angle between the velocity vector and the horizontal is $\theta = \arctan(\frac{gt}{\sqrt{2g(10-x)}})$ (downwards).

The angle between the inclined plane and the horizontal is 30° (upwards).

The condition for perpendicularity is that the angle between the velocity vector and the horizontal plus the angle between the inclined plane and the horizontal is 90°.

So, $\theta + 30^\circ = 90^\circ$, which means $\theta = 60^\circ$.

$\tan \theta = \tan 60^\circ = \sqrt{3}$.

$\frac{gt}{\sqrt{2g(10-x)}} = \sqrt{3}$.

$g^2 t^2 = 3 \cdot 2g(10-x)$.

$gt^2 = 6(10-x)$.

Now consider the position. Let the hole be at height x from the bottom. Let the point where the stream hits the inclined plane be at a horizontal distance X and vertical distance Y below the hole.

$X = \sqrt{2g(10-x)} t$

$Y = \frac{1}{2} g t^2$

The point (X, Y) is on the inclined plane.

Let's consider the similar question's approach. It uses the equation of the trajectory in terms of x and y.

$t = \frac{X}{\sqrt{2g(10-x)}}$.

$Y = \frac{1}{2} g (\frac{X}{\sqrt{2g(10-x)}})^2 = \frac{1}{2} g \frac{X^2}{2g(10-x)} = \frac{X^2}{4(10-x)}$.

The equation of the trajectory is $Y = \frac{X^2}{4(10-x)}$.

Now we need the equation of the inclined plane relative to the hole.

Let the inclined plane pass through a point $(X_p, Y_p)$ relative to the hole. The slope of the inclined plane is $-\tan(30^\circ)$ in the coordinate system with y downwards. The equation is $Y - Y_p = -\tan(30^\circ) (X - X_p)$. This is not helpful.

Let's consider the angle made by the position vector of the striking point relative to the hole with the horizontal. Let the striking point be P. Let the hole be H.

The coordinates of P are $(X, Y)$. The slope of the line HP is $\frac{Y}{X} = \frac{\frac{1}{2} g t^2}{\sqrt{2g(10-x)} t} = \frac{gt}{2\sqrt{2g(10-x)}}$.

We know $\frac{gt}{\sqrt{2g(10-x)}} = \sqrt{3}$.

So, the slope of HP is $\frac{\sqrt{3}}{2}$. The angle made by HP with the horizontal is $\beta$, where $\tan \beta = \frac{\sqrt{3}}{2}$.

Consider the triangle formed by the hole, the striking point on the inclined plane, and the horizontal line passing through the hole. Let the horizontal distance be X and the vertical distance be Y. The angle made by the inclined plane with the horizontal is 30°.

Let the angle of the position vector from the hole to the striking point with the horizontal be $\beta$.

The slope of the inclined plane is $\tan(30^\circ)$.

The stream strikes the plane normally. The velocity vector at the striking point makes an angle of 60° with the horizontal.

Let the striking point be at a distance R from the hole along some direction.

The coordinates of the striking point are $(R \cos \beta, R \sin \beta)$ relative to the hole, where $\beta$ is the angle below the horizontal.

So $X = R \cos \beta$ and $Y = R \sin \beta$.

The point $(X, Y)$ is on the inclined plane.

Let's use the relationship between the angle of the velocity vector and the angle of the position vector for a projectile.

For a projectile launched horizontally with velocity $v_0$, the trajectory is $y = \frac{g x^2}{2 v_0^2}$.

The slope of the tangent (velocity vector) at $(x, y)$ is $\frac{dy}{dx} = \frac{gx}{v_0^2}$.

The slope of the line joining the origin (launch point) to $(x, y)$ is $\frac{y}{x} = \frac{gx^2 / (2v_0^2)}{x} = \frac{gx}{2v_0^2}$.

The slope of the tangent is twice the slope of the line joining the origin to the point.

Let $\theta_v$ be the angle of the velocity vector with the horizontal, and $\theta_p$ be the angle of the position vector from the launch point with the horizontal.

$\tan \theta_v = 2 \tan \theta_p$.

In our case, the launch is from the hole. The velocity vector makes an angle of 60° below the horizontal when it strikes the plane. So $\theta_v = -60^\circ$.

The striking point is on the inclined plane. Let the angle of the position vector from the hole to the striking point with the horizontal be $\theta_p$.

$\tan(-60^\circ) = 2 \tan \theta_p$.

$-\sqrt{3} = 2 \tan \theta_p$.

$\tan \theta_p = -\frac{\sqrt{3}}{2}$.

This angle is measured from the horizontal. Since the striking point is below the hole and to the right, the angle should be negative. Let's consider the magnitude of the angle below the horizontal.

Let the angle of the position vector below the horizontal be $\beta$. $\tan \beta = \frac{\sqrt{3}}{2}$.

Now consider the geometry of the inclined plane. The inclined plane makes an angle of 30° with the horizontal.

Let the striking point be at a horizontal distance X and vertical distance Y below the hole.

The angle made by the line joining the hole to the striking point with the horizontal is $\beta$, where $\tan \beta = \frac{Y}{X} = \frac{\sqrt{3}}{2}$.

So, $Y = \frac{\sqrt{3}}{2} X$.

Now we need the equation of the inclined plane relative to the hole.

Let's assume the inclined plane starts at the horizontal level of the hole at a horizontal distance $X_0$. Then the equation of the inclined plane is $Y = \tan(30^\circ) (X - X_0)$. This cannot be correct as Y is positive downwards.

Let's assume the inclined plane starts at the bottom of the tank (height 0) at a horizontal distance $X_0$. The hole is at height x.

The equation of the inclined plane is $y_{bottom} = \tan(30^\circ) (x_{bottom} - X_0)$.

The coordinates of the striking point are $(X, x-Y)$ relative to the bottom.

So, $x-Y = \tan(30^\circ) (X - X_0)$.

$x - \frac{\sqrt{3}}{2} X = \frac{1}{\sqrt{3}} (X - X_0)$.

$\sqrt{3} x - \frac{3}{2} X = X - X_0$.

$X_0 = \frac{5}{2} X - \sqrt{3} x$.

We also have $gt^2 = 6(10-x)$.

$X = \sqrt{2g(10-x)} t$.

$X^2 = 2g(10-x) t^2 = 2g(10-x) \frac{6(10-x)}{g} = 12(10-x)^2$.

$X = \sqrt{12} (10-x) = 2\sqrt{3} (10-x)$.

$Y = \frac{\sqrt{3}}{2} X = \frac{\sqrt{3}}{2} (2\sqrt{3} (10-x)) = 3(10-x)$.

Now substitute into the equation of the plane:

$x - Y = \tan(30^\circ) (X - X_0)$.

$x - 3(10-x) = \frac{1}{\sqrt{3}} (2\sqrt{3}(10-x) - X_0)$.

$x - 30 + 3x = 2(10-x) - \frac{X_0}{\sqrt{3}}$.

$4x - 30 = 20 - 2x - \frac{X_0}{\sqrt{3}}$.

$6x - 50 = -\frac{X_0}{\sqrt{3}}$.

$X_0 = \sqrt{3} (50 - 6x)$.

Substitute $X_0 = \frac{5}{2} X - \sqrt{3} x$.

$\sqrt{3} (50 - 6x) = \frac{5}{2} (2\sqrt{3} (10-x)) - \sqrt{3} x$.

$\sqrt{3} (50 - 6x) = 5\sqrt{3} (10-x) - \sqrt{3} x$.

Divide by $\sqrt{3}$:

$50 - 6x = 5(10-x) - x$.

$50 - 6x = 50 - 5x - x$.

$50 - 6x = 50 - 6x$.

This equation is always true, which means the value of $X_0$ is consistent with the geometry, but it doesn't help us find x.

Let's look at the similar question's result $h = 8.33$ m.

In our problem, the height is x. So x should be 8.33 m.

Let's verify this. If x = 25/3 m, then $10-x = 10 - 25/3 = 5/3$ m.

$v_0 = \sqrt{2g(5/3)}$.

$gt^2 = 6(10-x) = 6(5/3) = 10$.

$t^2 = 10/g$. $t = \sqrt{10/g}$.

$\tan \theta = \frac{gt}{\sqrt{2g(10-x)}} = \frac{g\sqrt{10/g}}{\sqrt{2g(5/3)}} = \frac{\sqrt{10g}}{\sqrt{10g/3}} = \sqrt{3}$.

So $\theta = 60^\circ$. This confirms that if x = 25/3 m, the velocity vector makes an angle of 60° with the horizontal.

Now we need to check if the striking point is on the inclined plane.

$X = \sqrt{2g(10-x)} t = \sqrt{2g(5/3)} \sqrt{10/g} = \sqrt{10g/3} \sqrt{10/g} = \sqrt{100/3} = 10/\sqrt{3}$.

$Y = \frac{1}{2} gt^2 = \frac{1}{2} g (10/g) = 5$.

So, the striking point is at $(10/\sqrt{3}, 5)$ relative to the hole (x horizontal, y downwards).

The slope of the line joining the hole to the striking point is $\frac{Y}{X} = \frac{5}{10/\sqrt{3}} = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2}$.

$\tan \beta = \frac{\sqrt{3}}{2}$.

Now consider the inclined plane. Let the origin be at the bottom of the tank. The hole is at $(0, x)$.

The striking point is at $(X, x-Y) = (10/\sqrt{3}, 25/3 - 5) = (10/\sqrt{3}, 10/3)$.

The equation of the inclined plane is $y = \tan(30^\circ) (z - X_0)$.

The point $(10/\sqrt{3}, 10/3)$ is on the plane.

$10/3 = \frac{1}{\sqrt{3}} (10/\sqrt{3} - X_0)$.

$10/3 = 10/3 - \frac{X_0}{\sqrt{3}}$.

$\frac{X_0}{\sqrt{3}} = 0$. So $X_0 = 0$.

This means the inclined plane starts at the origin (bottom of the tank).

The equation of the inclined plane is $y = x \tan(30^\circ) = x/\sqrt{3}$ for $x \ge 0$.

The striking point is $(10/\sqrt{3}, 10/3)$.

Is $10/3 = (10/\sqrt{3}) / \sqrt{3}$?

$10/3 = 10/3$. Yes.

So, the assumption that the inclined plane starts at the bottom of the tank at horizontal position 0 is consistent with the result x = 25/3 m.

The value of x is 25/3 m.

We need to find the value of $0.6 \times x$.

$0.6 \times \frac{25}{3} = \frac{6}{10} \times \frac{25}{3} = \frac{6 \times 25}{10 \times 3} = \frac{150}{30} = 5$.