Question

A frog jumps vertically upward from a 20m tall building with an initial velocity of 8.1m/s. How high above the ground will the frog reach?

Answer 1

Answer:

Explanation:

Consider the initial position of the frog (20 m above ground) as the reference position. All measurements are positive measured upward.

Therefore,

u = 10 m/s, initial upward velocity.

H = - 20 m, position of the ground.

g = 9.8 m/s², acceleration due to gravity.

Part (a)

When the frog reaches a maximum height of h from the reference position, its velocity is zero. Therefore

u² - 2gh = 0

h = u²/(2g) = 10²/(2*9.8) = 5.102 m

At maximum height, the frog will be 20 + 5.102 = 25.102 m above ground.

Answer: 25.1 m above ground

Part (b)

Let v = the velocity when the frog hits the ground. Then

v² = u² - 2gH

v² = 10² - 2*9.8*(-20) = 492

v = 22.18 m/s

Answer: The frog hits the ground with a velocity of 22.2 m/s