Homework due ASAP physics 111

Tutor: None Selected Time limit: 1 Day

A blue ball is thrown upward with an initial speed of 21.2 m/s, from a height of 0.7 meters above the ground. 2.6 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 7.5 m/s from a height of 25.3 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.

3)What is the maximum height the blue ball reaches?


What is the height of the red ball 3.38 seconds after the blue ball is thrown? 


How long after the blue ball is thrown are the two balls in the air at the same height?
Sep 14th, 2015

Thank you for the opportunity to help you with your question!

3. maximum height

vf^2 - vi^2 = 2as

s = -vi^2/2a

s= -21.2m/s/2(-9.81m/s^2)

s= 1.081m

4 height of ball after 3.38 sec

time = 3.38 -2.6 = 0.78sec

d=vi *t + 1/2 at^2

d= 7.5m/s*0.78 + 1/2 * (9.81m/s^2 * 0.78^2)

d= 5.85m + 2.984 = 8.834m

height = 25.3 -8.843 = 16.466m

Please let me know if you need any clarification. I'm always happy to answer your questions.
Sep 14th, 2015

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