a cliff driver diver dove off a cliff at 10 m / s and landed in the water 48 m away from the cliff. How high is the nearest tenth]

Answer 1

You don't say so but I'm assuming the initial velocity was horizontal. This is a critical part of the problem but I'll go with that assumption.

In the horizontal direction, at 10 m/s the horizontal velocity doesn't change throughout the dive because the only force on the diver is the force of gravity acting in the y direction. So the time it takes the diver to go 48 m is (48 m)/(10m/s) = 4.8 s.

Now, looking at the y-direction, the distance y = v_iyt + 1/2at² but v_iy = 0 (based on the initial assumption that the diver dove straight out in the horizontal direction) so y = 1/2at² = 1/2(-9.81)(4.8)² = -113.0 m