Question

A ball thrown horizontally from a point 30.0 m above the ground strikes the ground after travelling a horizontal distance of 18.0 m. with what speed was it thrown? (g = 9.8 m/s²)

Answer 1

The components of the ball's position $r$ at time $t$ are

$\begin{cases}r_x(t)=v_{0x}t\\v_y(t)=30.0-4.9t^2\end{cases}$

The ball stops 18.0 m from where it began, so that

$\begin{cases}18.0=v_{0x}t\\0=30.0-4.9t^2\end{cases}$

From the second equation, we can show that the ball travels for about $t=2.47$ seconds, which means it was initially thrown with a horizontal velocity of

$18.0\,\mathrm m=v_{0x}(2.47\,\mathrm s)\implies v_{0x}=7.29\,\dfrac{\mathrm m}{\mathrm s}$