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maiad
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Homework Statement
A blue ball is thrown upward with an initial speed of 20.4 m/s, from a height of 0.5 meters above the ground. 2.5 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 9.8 m/s from a height of 23.9 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.
How long after the blue ball is thrown are the two balls in the air at the same height
Homework Equations
Xf=Xi+Vit+0.5a^2
The Attempt at a Solution
Blue ball:
Vi=20.4 m/s
a=-9.81
Xi=0
t=t-2.5
Red ball:
Vi=-9.8
a=-9.81
xi=23.4
t=t
The height is different because i chose the blue ball as reference point.
I plugged the values into the equation and equated them giving me:
20.4(t-2.5)-4.9(t+2.5)^2=23.4-9.8t-4.9t^2
further simplifying i get...
-4.1t-4.9t^2-81.6=23.4-9.8t-4.9t^2
then...
5.7t=105
t=18.4s
which is wrong. can someone tell me where i went wrong?
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