Question

A blue ball is thrown upward with an initial speed of 21.4 m/s, from a height of 0.9 meters above the ground. 2.6 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 9.2 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.

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

Answer 1

let after t time the two are at same height

for blue ball

y1-0.9 = v1y*t + 0.5*g*t^2

y1 - 0.9 = (21.4*t) - (0.5*9.8*t^2)

y1 - 0.9 = 21.4t - 4.9t^2

y1 = 0.9 + 21.4t - 4.9t^2.............(1)

for red ball

time = t-2.6

25.3 - y1 = v2*(t-2.6) + 0.5*g*(t-2.6)^2

25.3 - y1 = 9.2*(t-2.6) + 4.9*(t^2 + 6.76 - 5.2t)

25.3 - y1 = 9.2t - 23.92 + 4.9t^2 + 33.124 - 25.48t

25.3 - y1 = 4.9t^2 - 16.28t + 9.204

y1 = 16.096 - 4.9t^2 + 16.28t .........(2)

0.9 + 21.4t - 4.9t^2 = 16.096 - 4.9t^2 + 16.28t

(21.4-16.28)*t = 16.096-0.9

t = 2.967 s <--------answer