Posted by **Anonymous** on Monday, September 17, 2012 at 4:31pm.

A projectile is fired from the top of a cliff at 14.2 m/s at an angle of 36.9 degrees below the horizontal, and strikes water at the same level as the base of the cliff 3.51 seconds later. (Assume g=10.0 m/s2.)

How high was the top of the cliff above the water?

- Physics 124 -
**Steve**, Monday, September 17, 2012 at 5:14pm
all we care about is the downward velocity. Initially that is

14.2 cos -36.9° = -11.36 m/s

so, starting from height H, the height after t seconds is

-11.36t - 4.9 t^2

at t=3.51,

h = -100.24 m

So, the cliff is 100.24m high

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