A red ball is thrown down with an initial speed of 1.3 m/s from a height of 26 m

Question

A red ball is thrown down with an initial speed of 1.3 m/s from a height of 26 meters above the ground. Then, 0.5 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 24.4 m/s, from a height of 0.7 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.

Question 1 ) How long after the red ball is thrown are the two balls in the air at the same height???? please focuse on the question

Explanation / Answer

g=9.81m/s^2
red ball: h=h0-v0t-1/2gt^2
h= 26-1.3t-4.905t^2
blue ball: h=h0+v0t-1/2gt^2
h= 0.7+24.4(t-0.5)-4.905(t-0.5)^2= 29.305t-12.72625-4.905t^2
h=h
26-1.3t = 29.305t-12.72625
30.605t= 38.72625
t= 1.265 s

red: h= 26-1.3*1.265-4.9*1.265^2= 16.5 m
blue: h= 0.7+24.4*0.765-4.9*0.765^2= 16.5 m

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