Question

person throws a ball horizontally from the top of a building that is 24.0 m above the ground level. The ball lands 100 m down range from the base of the building. What was the initial velocity of the ball? Neglect air resistance and use g = 9.81 m/s2.

Answer 1

Answer:45.24 m/s

Explanation:

Given

Height of Building h=24 m

Range of ball R=100 m

Considering Vertical motion of ball

using $y=u_yt+\frac{a_yt^2}{2}$

initial vertical velocity is zero therefore $u_y=0$

$24=0+\frac{9.8\times t^2}{2}$

$t=\sqrt{\frac{48}{9.8}}$

$t=2.21 s$

Now considering Horizontal Motion

$R=u_xt+\frac{a_xt^2}{2}$

$100=u_x\times 2.21+0$  , as there is no horizontal acceleration

$u_x=45.24 m/s$