Posted by **Emily** on Tuesday, October 11, 2011 at 8:12pm.

A person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of vi = 13 m/s. The cliff is h = 58 m above the water's surface, as shown below.

(a) How long does it take for the stone to fall to the water?

s

(b) With what speed does it strike the water?

m/s

- physics -
**bobpursley**, Tuesday, October 11, 2011 at 8:20pm
h=1/2 g t^2 solve for time t to hit the surface.

speed? ?

in the vertical v= at

in the horizontal v=13m/s

vf= sqrt(vvertical^2+13^2)

- physics -
**drwls**, Tuesday, October 11, 2011 at 8:24pm
Assume the initial kick is in a horizontal direction. Vertical motion is not affected.

(a)

(1/2) g t^2 = 58 m when it hits the water

Solve for t.

(b) The new kinetic energy will be the initial kinetic energy PLUS M g H

where H = 58 m.

(M/2)*Vfinal^2 = (M/2)*vi^2 + M g H

Cancel the M's and solve for Vfinal

Vfinal^2 = vi^2 + 2 g H

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