A blue ball is thrown upward with an initial speed of 22.2 m/s, from a height of

Question

A blue ball is thrown upward with an initial speed of 22.2 m/s, from a height of 0.6 meters above the ground. 2.7 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 5.9 m/s from a height of 26.7 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2 1) What is the speed of the blue ball when it reaches its maximum height? m/s Submit 2) How long does it take the blue ball to reach its maximum height? S Submit 3) What is the maximum height the blue ball reaches? m Submit 4) What is the height of the red ball 3.51 seconds after the blue ball is thrown? m Submit 5) How long after the blue ball is thrown are the two balls in the air at the same height? S Submit

Explanation / Answer

1)

At maximum height, speed is always 0

Answer: 0 m/s

2)

vi = 22.2 m/s

a = -9.81 m/s^2

vf = 0 m/s

t = ?

use:

vf = vi + a*t

0 = 22.2 - 9.81*t

t = 22.2/9.81

= 2.26 s

Answer: 2.26 s

3)

vi = 22.2 m/s

a = -9.81 m/s^2

vf = 0 m/s

1st find the maximum distance from where its thrown:

vf^2 = vi^2 + 2*a*d

0 = 22.2^2 + 2*(-9.81)*d

d = 492.84/19.62

= 25.1 m

It was already thrown from 0.6 m

so,

max height = 25.1 + 0.6 = 25.7 m

Answer: 25.7 m

4)

time taken by red ball,

t = 3.51 - 2.7 = 0.81 s

vi = -5.9 m/s

a = -9.81 m/s^2

find the displacement of blue ball:

d = vi*t + 0.5*a*t^2

= -5.9*0.81 + 0.5*(-9.81)*0.81^2

= -4.779 - 3.22

= - 8.0 m

It was thrown from 26.7 m

So, now its height = 26.7 - 8.0 = 18.7 m

Answer: 18.7 m

Only 4 parts at a time

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