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

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

Let the projectile's initial velocity be $u$.

Time of flight $T=3$ s (given)

Using $T=\sqrt{\frac{2h}{g}}$ $⟹h=\frac{g{T}^2}{2}$

$\therefore$ $h=\frac{9.8\times 3^2}{2}=44.1$ m

x direction : $a_x=0$ $⟹V_x=u$

y direction : $V_y=u_y+a_{y}T$

$\therefore$ $V_y=0+9.8\times 3=29.4m/s$

Also $\frac{V_y}{V_x}=tan{45}^o=1$

$⟹$ $V_x=V_y=29.4$ m/s

Thus initial speed of the projectile $u=V_x=29.4m/s$