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August 30, 2014

August 30, 2014

Posted by **mb** on Friday, September 26, 2008 at 9:44am.

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0m above a flat, horizontal beach. With what angle of impact does the stone land (in degrees).

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18 m/s. the cliff is 50 m above a flat, horizontally beach. how long after being released does the stone strike the beach below the cliff?

A student atands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0 m above flat, horizontal beach. With what speed of impact does the stone land?

- physics -
**drwls**, Friday, September 26, 2008 at 11:07amThe horizontal velocity component remains

Vx = 18.0 m/s

during the fall to the beach.

The vertical component of velocity (measured positive upwards) is

Vy = - gt

The time t that it takes to reach the beach when falling from height H is given by solving

H = (g/2) t^2

Therefore t = sqrt (2H/g) = 3.19 s

That answers your second question. Use that value of t to compute Vy at impact and from that and Vx get the angle of impact.

The speed of impact is sqrt[Vx^2 + Vy^2]

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