Question

A stone is projected from horizontal ground with minimum initial velocity $v_0$ m/s such that the stone is able to cross a sloped roof $AB$ as shown below. Find the value of $v_0$.

Answer 1

14.56

Answer 2

The problem asks for the minimum initial velocity $v_0$ required for a stone projected from the horizontal ground to clear a sloped roof AB. The roof has point A at height 10 m and point B at height 5 m. The length of the roof AB is 25 m.

1. Determine the geometry of the roof:

   Let the horizontal distance between points A and B be $\Delta x$. The vertical height difference between A and B is $\Delta y = 10 \text{ m} - 5 \text{ m} = 5 \text{ m}$. Using the Pythagorean theorem for the length of the roof: $L^2 = (\Delta x)^2 + (\Delta y)^2$ $25^2 = (\Delta x)^2 + 5^2$ $625 = (\Delta x)^2 + 25$ $(\Delta x)^2 = 600$ $\Delta x = \sqrt{600} = 10\sqrt{6} \text{ m}$.

2. Formulate the problem using the trajectory equation:

   Let the stone be projected from the origin (0,0) with initial velocity $v_0$ at an angle $\theta$ with the horizontal. The equation of the trajectory is: $y = x \tan\theta - \frac{g x^2}{2 v_0^2 \cos^2\theta}$ This can be rewritten as: $y = x \tan\theta - \frac{g x^2}{2 v_0^2} (1 + \tan^2\theta)$ Rearranging to express $v_0^2$: $v_0^2 = \frac{g x^2 (1 + \tan^2\theta)}{2 (x \tan\theta - y)}$

3. Apply the condition for minimum velocity to clear a line segment:

   For the stone to clear the roof AB with minimum initial velocity, the trajectory must be tangent to the line segment AB. This means the parabola of the projectile path touches the line segment AB at exactly one point. Let the coordinates of point A be $(x_1, y_1)$ and point B be $(x_2, y_2)$. We have $y_1 = 10$ m and $y_2 = 5$ m. Let $x_1$ be the horizontal distance from the origin to point A. Then $x_2 = x_1 + \Delta x = x_1 + 10\sqrt{6}$.

   Consider the equation of trajectory: $y = x \tan\theta - \frac{g x^2}{2 v_0^2}(1+\tan^2\theta)$. Let $P = \tan\theta$. $y = xP - \frac{g x^2}{2 v_0^2}(1+P^2)$ $2v_0^2 y = 2v_0^2 xP - gx^2(1+P^2)$ $gx^2 P^2 - 2v_0^2 xP + (gx^2 + 2v_0^2 y) = 0$

   For a given $(x,y)$, this quadratic in $P$ has real solutions for $P$ (i.e., real $\theta$) if the discriminant is non-negative: $D = (-2v_0^2 x)^2 - 4(gx^2)(gx^2 + 2v_0^2 y) \ge 0$ $4v_0^4 x^2 - 4g^2 x^4 - 8g v_0^2 x^2 y \ge 0$ Divide by $4x^2$ (assuming $x \neq 0$): $v_0^4 - g^2 x^2 - 2g v_0^2 y \ge 0$ $v_0^4 - 2gy v_0^2 - g^2 x^2 \ge 0$ The minimum $v_0^2$ to reach a point $(x,y)$ occurs when the discriminant is zero: $v_0^2 = \frac{2gy \pm \sqrt{4g^2 y^2 + 4g^2 x^2}}{2} = gy \pm g\sqrt{x^2+y^2}$. Since $v_0^2 > 0$, we take the positive root: $v_0^2 = g(y + \sqrt{x^2+y^2})$.

   The problem requires the trajectory to clear the entire line segment AB. This means all points $(x,y)$ on the line segment must be reachable. The minimum velocity $v_0$ is such that the trajectory is tangent to the line segment AB. Let the line segment be represented by $y = mx+c$. The equation for the minimum velocity $v_0$ for a projectile launched from the origin to just touch a line $y = mx+c$ is: $v_0^2 = \frac{g c}{1 - 2m \tan\theta + m^2}$ where $\tan\theta = \frac{m + \sqrt{m^2-1+c/H}}{1}$

   Let's use the concept of the parabola of safety. The minimum velocity to clear the line segment occurs when the line segment is tangent to the parabola of safety. The equation of the parabola of safety (envelope of trajectories) is $y = \frac{v_0^2}{2g} - \frac{g x^2}{2v_0^2}$. Let the line segment AB be $y = mx+c$. We need to find $v_0$ such that $mx+c = \frac{v_0^2}{2g} - \frac{g x^2}{2v_0^2}$ has exactly one solution for $x$. $\frac{g x^2}{2v_0^2} + mx + (c - \frac{v_0^2}{2g}) = 0$. For a unique solution, the discriminant must be zero: $m^2 - 4 \left(\frac{g}{2v_0^2}\right) \left(c - \frac{v_0^2}{2g}\right) = 0$ $m^2 - \frac{2g}{v_0^2} (c - \frac{v_0^2}{2g}) = 0$ $m^2 - \frac{2gc}{v_0^2} + 1 = 0$ $v_0^2 (m^2+1) = 2gc$ $v_0^2 = \frac{2gc}{m^2+1}$

   Now we need to find $m$ and $c$ for the line segment AB. $m = \frac{-1}{2\sqrt{6}}$. The equation of the line AB is $y - 10 = \frac{-1}{2\sqrt{6}}(x - x_A)$. $y = \frac{-1}{2\sqrt{6}}x + \left(\frac{x_A}{2\sqrt{6}} + 10\right)$. So, $c = \frac{x_A}{2\sqrt{6}} + 10$. We need to minimize $v_0^2 = \frac{2g(\frac{x_A}{2\sqrt{6}} + 10)}{(\frac{-1}{2\sqrt{6}})^2 + 1} = \frac{2g(\frac{x_A}{2\sqrt{6}} + 10)}{\frac{1}{24} + 1} = \frac{2g(\frac{x_A}{2\sqrt{6}} + 10)}{\frac{25}{24}}$. $v_0^2 = \frac{48g}{25} \left(\frac{x_A}{2\sqrt{6}} + 10\right)$. To minimize $v_0^2$, we must minimize $x_A$. The line segment AB extends from $x=x_A$ to $x=x_A+10\sqrt{6}$. The point of tangency $(x_t, y_t)$ must lie on the line segment AB, i.e., $x_A \le x_t \le x_A+10\sqrt{6}$. The x-coordinate of the point of tangency is given by: $x_t = - \frac{m v_0^2}{g} = - \frac{m}{g} \frac{2gc}{m^2+1} = - \frac{2mc}{m^2+1}$. Substitute $m = \frac{-1}{2\sqrt{6}}$ and $c = \frac{x_A}{2\sqrt{6}} + 10$: $x_t = - \frac{2(\frac{-1}{2\sqrt{6}})(\frac{x_A}{2\sqrt{6}} + 10)}{(\frac{-1}{2\sqrt{6}})^2 + 1} = \frac{\frac{1}{\sqrt{6}}(\frac{x_A}{2\sqrt{6}} + 10)}{\frac{1}{24} + 1} = \frac{\frac{x_A}{12} + \frac{10}{\sqrt{6}}}{\frac{25}{24}}$ $x_t = \frac{24}{25} \left(\frac{x_A}{12} + \frac{10}{\sqrt{6}}\right) = \frac{2x_A}{25} + \frac{240}{25\sqrt{6}} = \frac{2x_A}{25} + \frac{48\sqrt{6}}{25}$.

   We need $x_A \le x_t \le x_A + 10\sqrt{6}$. For minimum $v_0$, the tangent point must be the leftmost point A, i.e., $x_t = x_A$. This means the trajectory is tangent to the line AB at point A. So, $x_A = \frac{2x_A}{25} + \frac{48\sqrt{6}}{25}$. $25x_A = 2x_A + 48\sqrt{6}$ $23x_A = 48\sqrt{6}$ $x_A = \frac{48\sqrt{6}}{23}$.

   Now substitute this $x_A$ back into the $v_0^2$ equation: $v_0^2 = \frac{48g}{25} \left(\frac{\frac{48\sqrt{6}}{23}}{2\sqrt{6}} + 10\right) = \frac{48g}{25} \left(\frac{48}{46} + 10\right) = \frac{48g}{25} \left(\frac{24}{23} + 10\right)$ $v_0^2 = \frac{48g}{25} \left(\frac{24 + 230}{23}\right) = \frac{48g}{25} \left(\frac{254}{23}\right)$ $v_0^2 = \frac{12192g}{575}$.

   Let's use $g=10 \text{ m/s}^2$. $v_0^2 = \frac{12192 \times 10}{575} = \frac{121920}{575} = \frac{24384}{115}$. $v_0 = \sqrt{\frac{24384}{115}} \approx \sqrt{211.9} \approx 14.55 \text{ m/s}$.

   This value seems reasonable. Let's recheck the assumption that $x_t=x_A$. This is true if the minimum of $v_0$ occurs when the tangent point is at the start of the line segment.

Final calculation: $v_0^2 = \frac{48g}{25} \left(\frac{x_A}{2\sqrt{6}} + 10\right)$ $x_A = \frac{48\sqrt{6}}{23}$ $v_0^2 = \frac{48g}{25} \left(\frac{48\sqrt{6}}{23 \times 2\sqrt{6}} + 10\right) = \frac{48g}{25} \left(\frac{24}{23} + 10\right)$ $v_0^2 = \frac{48g}{25} \left(\frac{24 + 230}{23}\right) = \frac{48g}{25} \left(\frac{254}{23}\right)$ $v_0^2 = \frac{12192g}{575}$ Using $g=9.8 \text{ m/s}^2$: $v_0^2 = \frac{12192 \times 9.8}{575} = \frac{119481.6}{575} \approx 207.808$ $v_0 = \sqrt{207.808} \approx 14.415 \text{ m/s}$.

Using $g=10 \text{ m/s}^2$: $v_0^2 = \frac{12192 \times 10}{575} = \frac{121920}{575} = \frac{24384}{115} \approx 211.99$ $v_0 = \sqrt{211.99} \approx 14.56 \text{ m/s}$.

Let's verify the condition $x_A \le x_t \le x_A+10\sqrt{6}$. We set $x_t = x_A$. So $x_A = \frac{48\sqrt{6}}{23} \approx \frac{48 \times 2.449}{23} \approx \frac{117.55}{23} \approx 5.11 \text{ m}$. The horizontal extent of the roof is $10\sqrt{6} \approx 10 \times 2.449 = 24.49 \text{ m}$. So $x_A+10\sqrt{6} \approx 5.11 + 24.49 = 29.6 \text{ m}$. Since $x_t = x_A$, the tangency point is at the beginning of the roof. This means the trajectory just touches point A. This is a valid condition for minimum velocity.