Question

A projectile is projected from point 'O' inside a vertical triangle ABC with base angles 30° and 60°. Projectile is projected in such a way that it touches at points P and Q of triangle as shown in figure. Take horizontal speed of projectile as $\sqrt{3}m/s$.

The velocity at point 'P' is

• A. 1 m/s
• B. 2 m/s
• C. 3 m/s
• D. 4 m/s

Answer 1

Answer: (B)

2 m/s

Answer 2

The problem asks for the velocity of the projectile at point P.

1. Understand the projectile motion properties:

   • The horizontal component of the projectile's velocity ($v_x$) remains constant throughout the motion because there is no horizontal acceleration (neglecting air resistance). We are given $v_x = \sqrt{3} \text{ m/s}$.
   • At any point on the trajectory, the velocity vector is tangent to the path.

2. Analyze the velocity at point P:

   • Point P is where the projectile's path touches the side AB of the triangle. This means the side AB is tangent to the projectile's trajectory at P.
   • The angle that side AB makes with the horizontal base AC is given as 30°. Therefore, the velocity vector at point P makes an angle of 30° with the horizontal.
   • Let the velocity at P be $\vec{v}_P$. Its horizontal component is $v_{xP}$ and its vertical component is $v_{yP}$.

3. Calculate the components of velocity at P:

   • Horizontal component: $v_{xP} = v_x = \sqrt{3} \text{ m/s}$.
   • Vertical component: The tangent of the angle the velocity vector makes with the horizontal is the ratio of the vertical component to the horizontal component. $\tan(30^\circ) = \frac{v_{yP}}{v_{xP}}$ We know $\tan(30^\circ) = \frac{1}{\sqrt{3}}$. So, $\frac{1}{\sqrt{3}} = \frac{v_{yP}}{\sqrt{3} \text{ m/s}}$. Solving for $v_{yP}$: $v_{yP} = \sqrt{3} \times \frac{1}{\sqrt{3}} = 1 \text{ m/s}$.

4. Determine the velocity at P:

   The velocity at point P can be expressed as a vector: $\vec{v}_P = v_{xP} \hat{i} + v_{yP} \hat{j} = \sqrt{3} \hat{i} + 1 \hat{j} \text{ m/s}$.

   The magnitude of the velocity at P (speed at P) is: $v_P = \sqrt{v_{xP}^2 + v_{yP}^2}$ $v_P = \sqrt{(\sqrt{3})^2 + (1)^2}$ $v_P = \sqrt{3 + 1}$ $v_P = \sqrt{4}$ $v_P = 2 \text{ m/s}$.

   The direction of the velocity at P is 30° above the horizontal.

Since the question asks for "The velocity at point 'P'" without specifying "speed" or "components", it typically implies the magnitude of the velocity in such contexts.