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

hotter stars = bluer stars

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.