Description
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?
4)
What is the height of the red ball 3.38 seconds after the blue ball is thrown?5)
How long after the blue ball is thrown are the two balls in the air at the same height?Explanation & Answer
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
Review
Review
24/7 Homework Help
Stuck on a homework question? Our verified tutors can answer all questions, from basic math to advanced rocket science!