Question

A ball is launched horizontally from a height of 19.6 m above the ground and it strikes the ground after moving 10.0 m horizontally. What was the initial speed of the ball?

Answer 1

Since the ball is launched horizontally, the vertical component of the velocity will be zero.

Now for vertical direction(Y-direction),

using the kinematical equation,

$s=ut+\frac{1}{2}at^2$

since the vertical component of initial-velocity is zero, $u=0$ . Also we have been given, $s=19.6\,m$ . The acceleration in this case is the gravitational acceleration

$\therefore 19.6=\frac{1}{2}\cdot 9.8\cdot t^2$

$\therefore t^2=4 \\\therefore t=2\,s$

(we have ignored the negative root t=-2s because time cannot be negative)

Now the horizontal component of velocity is unaffected because there is no acceleration in horizontal direction or X-direction. So the initial horizontal velocity is same as final horizontal velocity. The ball travels 10 m of distance in 2 seconds. Therefore,

$\boxed{u=v=\frac{10}{2}=5\,m/s}$

Since there is no vertical component of velocity initially, this is the initial speed of the ball.