Help with projectile motion...?

Question

A diver running 1.6m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 3.0s later. How high was the cliff?

The answer key says 4.8m, but I don't understand how they got the answer.

Answer 1

Since his horizontal velocity component is constant: 1.6 m/s

His horizontal distance from the cliff is:

Sh=v*t=1.6 m/s * 3 s

=4.8 meters

the height of the cliff is: s=0.5*g*t^2

Sv=0.5 * 9.8 * (3^2) = 44.1 meters

hope this helps.

Answer 2

Are you sure it was not how far away from the cliff the diver hits the water? Cuz, then you just use

Vx*t = 1.6m/s * 3s = 4.8 m away horizontally from the cliff

horizontal motion does not affect vertical motion: so height of cliff should be( which ithink is the answer you may have got originally):

y = 0.5 a t^2 = 0.5 (-9.8) (3s)^2 = 44.1 m

Hope that helps

Answer 3

1.6m/s * 3.0 s = 4.8 m