Asked by Sandhya
A clump of soft clay is thrown horizontally from 78.40 m above the ground with a speed of 27.0 m/s. Where is the clay after 4.5 s? Assume it sticks in place when it hits the ground.
horizontal ....m from the launch position
vertical ....m from the ground
horizontal ....m from the launch position
vertical ....m from the ground
Answers
Answered by
drwls
With no initial vertical velocity component, it hits the ground after time T given by
78.4 = (g/2) T^2
T = 4.00 s
It's horizontal location is the initial (horizontal) velocity muliplied by that time. It does not move after that.
78.4 = (g/2) T^2
T = 4.00 s
It's horizontal location is the initial (horizontal) velocity muliplied by that time. It does not move after that.
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