Question

A body projected horizontally with 98 $m{s}^{-1}$ from the top of a tower reaches the ground in 10s. Then it strikes the ground at angle:

• A. $0^0$
• B. ${30}^0$
• C. ${45}^0$
• D. ${60}^0$

Answer 1

The correct option is C ${45}^0$

Given,

Vertical velocity is zero

Horizontal velocity, $v_x=98m{s}^{-1}$

time, $t=10sec$

Apply kinematic equation of motion

$v_y=u_y+gt$

$v_y=0+9.8\times 10=98m{s}^{-1}$

Vertical velocity, $v_y=98m{s}^{-1}$

$\theta ={\tan }^{-1}\left(\frac{v_y}{v_x}\right)={\tan }^{-1}\left(\frac{98}{98}\right)$

$\theta ={45}^o$

Hence, angle is ${45}^o$