A body is thrown horizontally from the top of a tower and strikes the ground after three seconds at an angle of ${45}^{\circ }$ with the horizontal. Find the height of the tower and the speed with which the body was projected. Take $g=9.8m/s^2$

The correct option is A. 29.4 m/s.

Given,

$u_y=0$ and $a_y=g=9.8m/s^2$

Using the first equation of motion we get

$s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$s_y=0\times 3+\frac{1}{2}\times 9.8\times 3^2$

$s_y=44.1m$

Therefore, the height of the tower is 44.1 m.

From the first equation of motion

$v_y=u_y+a_{y}t=0+9.8\left(3\right)$

$v_y=29.4m/s$

As the resultant velocity v makes an angle of ${45}^{\circ }$ with the horizontal, so

$tan{45}^{\circ }=\frac{v_y}{v_x}$ or $1=\frac{29.4}{v_x}$

$v_x=29.4m/s$

Therefore, the speed with which the body was projected (horizontally) is 29.4 m/s.