Question

A body is projected from the surface of the earth with a velocity of $10 \sqrt{3} \, \text{m/s}$ such that its range is maximum. The velocity of the body at half of the maximum height is:

• A. $10 \sqrt{3} \, \text{ms}^{-1}$
• B. $15 \, \text{ms}^{-1}$
• C. $15 \sqrt{2} \, \text{ms}^{-1}$
• D. $30 \, \text{ms}^{-1}$

Hint:

The velocity at half the maximum height involves only reduced vertical velocity while horizontal remains constant, calculated using the energy conservation principle.

Answer 1

The Correct Option is B

$$\text{Given initial velocity, } u = 10 \sqrt{3} \, \text{m/s} \text{ at } 45^\circ \text{, }$$ $$u_x = u_y = 5 \sqrt{6} \, \text{m/s}. \text{ Time to max height, } t_{\text{max}} = \frac{5 \sqrt{6}}{9.8}$$ $$H = u_y t_{\text{max}} - \frac{1}{2} g t_{\text{max}}^2$$ $$v_y \text{ at } \frac{H}{2} = \sqrt{u_y^2 - g \left(u_y \frac{t_{\text{max}}}{2}\right)} = 5 \sqrt{3} \, \text{m/s}$$ $$\text{Total velocity at half max height, } v = \sqrt{u_x^2 + v_y^2} = 15 \, \text{m/s}$$