A diver running 2.2 m/s drives out horizontally from the edge of a vertical cliff and 2.7 s later reaches the water below.

Question

Part A How high was the cliff ?

Express your answer using two significant figures.

Part B

How far from its base did the diver hit the water?

Express your answer using two significant figures.

Answer 1

1. Using Δy = V_iyt + 1/2a_yt² compute the height of the cliff by finding delta y as follows:

Separate the x and y directions: initial velocity in the y-direction is 0, Acceleration in the y-direction is gravity = -9.81 m/s², time falling = 2.7s

Δy = 0*2.7s + 1/2 (-9.8m/s²)*(2.7s)² = -35.75 m

2 Because the diver is traveling in the x-direction at constant velocity of 2.2 m/s throughout the duration in the air, and there is NO acceleration in the x-direction. Use Δx = V_ixt + 1/2a_xt²

Δx = 2.2 m/s *2.7s = 5.9 m