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Natalija [7]
3 years ago
8

A ball is thrown straight up from the ground with an unknown velocity. It reaches its highest point after 3.5s with what velocit

y did it leave the ground
Physics
1 answer:
Evgen [1.6K]3 years ago
7 0

The ball's velocity at time t would be given by

v_y=v_{0y}-gt

At its highest point, we'll have v_y=0. We're told this occurs at t=3.5\,\mathrm s, which means

0=v_{0y}-\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)(3.5\,\mathrm s)

\implies v_{0y}=34\,\dfrac{\mathrm m}{\mathrm s}

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