A cliff diver from the top of a 100 (m) cliff. He begins his dive by jumping up with a velocity of 5 (m/s) a. How long does it take for him to hit the water below? b. What is his velocity right before he hits the water?

A.) 4.81 seconds

B.) 44.6 m/s

Explanation:

He begins his dive by jumping up with a velocity of 5 (m/s).

Let us first calculate the maximum height reached by using third equation of motion

V^2 = U^2 - 2gH

At maximum height, V = 0

0 = 5^5 - 2 × 9.8H

19.6H = 25

H = 25 /19.6

H = 1.28 m

The time taken for the diver to reach the water from the maximum height can be calculated by using second equation of motion.

Where height h = 1.28 + 100 = 101.28 m

h = Ut + 1/2gt^2

As the diver drop from maximum height, U = 0

101.28 = 1/2 × 9.8 × t^2

4.9t^2 = 101.28

t^2 = 101.28/4.9

t^2 = 20.669

t = sqrt ( 20.669)

t = 4.55s

As the diver jumped up, the time taken to reach the maximum height will be

Time = 1.28 / 5 = 0.256

The time taken for him to hit the water below will be 0.256 + 4.55 = 4.81 seconds

B.) Velocity right before he hits the water will be

V^2 = U^2 + 2gH

But U = 0

V^2 = 2 × 9.8 × 101.28

V^2 = 1985.09

V = 44.6 m/s

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures