A blue ball is thrown upward with an initial speed of 23.9 m/s, from a height of

Question

A blue ball is thrown upward with an initial speed of 23.9 m/s, from a height of 0.8 meters above the ground.   2.9 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 10.7 m/s from a height   of  30.9 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.

What is the height of the red ball 3.77 seconds after the blue ball is thrown? ANSWER IS 17.9

How long after the blue ball is thrown are the two balls in the air at the same height?

PLEASE SHOW STEPS

Explanation / Answer

missed that the blue ball was thrown down

a) yred = 30.9 -10.7*(3.77-2.9) - 0.5*9.81*(3.77-2.9)^2=
17.88 m

b) y blue = 0.8 + 23.9*t - 0.5*9.81*t^2

y red = 30.9 - 10.7*(t-2.9) - 0.5*9.81*(t-2.9)^2

y blue = yred

0.8 + 23.9*t - 0.5*9.81*t^2=30.9 - 10.7*(t-2.9) - 0.5*9.81*(t-2.9)^2

t=3.23 s

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