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yaroslaw [1]
3 years ago
9

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)
Physics
1 answer:
AURORKA [14]3 years ago
7 0

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}

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