Asked by Abdu
The fist ball is thrown up ward with avelocity of 20m/s at the same instant. The second ball is dropped at 40m hight from the first ball at what time the two ball collide each other?
Answers
Answered by
drwls
They collide when Y1 = Y2. Those are the heights of the two balls, as measured from the a defined "zero" height location.
Y1 = 20 t - 4.9 t^2
Y2 = 40 - 4.9 t^2
When Y1 = Y2,
20 t = 40
t = 2 seconds
Y = 40 - 4.9 t^2 = 20.4 m above the ground
Y1 = 20 t - 4.9 t^2
Y2 = 40 - 4.9 t^2
When Y1 = Y2,
20 t = 40
t = 2 seconds
Y = 40 - 4.9 t^2 = 20.4 m above the ground
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