Question

The kinetic energy at the highest point of the trajectory of a projectile is $200 \text{ J}$. If the mass of the projectile is $1 \text{ Kg}$ and the maximum height reached by it is $20 \text{ m}$, then velocity of the projectile from the ground is

• A. $20 \text{ m/s}$
• B. $10 \text{ m/s}$
• C. $20\sqrt{2} \text{ m/s}$
• D. $10\sqrt{2} \text{ m/s}$

Hint:

In projectile motion, horizontal velocity never changes (neglecting air resistance). If you find the horizontal speed at the top, you've found the horizontal speed at the launch point.

Answer 1

The Correct Option is C

To find the initial velocity ($u$), we must determine its horizontal component ($u_x$) and vertical component ($u_y$) using the given kinetic energy at the peak and the maximum height. Step 1: Find the Horizontal Velocity ($u_x$) At the highest point of a trajectory, the vertical velocity is zero. Only the horizontal component remains, which is constant throughout the flight ($v_{top} = u \cos \theta$). Given Kinetic Energy at top ($KE_{top}$) = $200 \text{ J}$ and $m = 1 \text{ kg}$: $$KE_{top} = \frac{1}{2} m (u \cos \theta)^2$$ $$200 = \frac{1}{2} (1) (u_x)^2 \implies u_x^2 = 400 \implies u_x = 20 \text{ m/s}$$

Step 2: Find the Initial Vertical Velocity ($u_y$) The maximum height ($H$) formula is $H = \frac{u_y^2}{2g}$. Taking $g = 10 \text{ m/s}^2$: $$20 = \frac{u_y^2}{2(10)} \implies 20 = \frac{u_y^2}{20}$$ $$u_y^2 = 400 \implies u_y = 20 \text{ m/s}$$

Step 3: Calculate Resultant Initial Velocity ($u$) The total velocity from the ground is the vector sum of these components: $$u = \sqrt{u_x^2 + u_y^2}$$ $$u = \sqrt{20^2 + 20^2} = \sqrt{400 + 400} = \sqrt{800}$$ $$u = 20\sqrt{2} \text{ m/s}$$