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\,(\mathrm m)/(\mathrm s)`