Asked by roger
a stone is thrown straight upward and rises to a maximum vertical height of 40.0m. with what velocity was it thrown and how long does it take for the stone to return to the initial level from which it was thrown?
Answers
Answered by
MathMate
Ignore air resistance => no energy loss
PE at 40.0 m = mgh = 40*9.8m J
=392 J
PE=KE=(1/2)mv²
vi=sqrt(2gh)=sqrt(2*9.8*40)
= 28 m/s
Time = 2 * time it takes to reach highest point
= 2*(28/9.8)
= 5.71 s
PE at 40.0 m = mgh = 40*9.8m J
=392 J
PE=KE=(1/2)mv²
vi=sqrt(2gh)=sqrt(2*9.8*40)
= 28 m/s
Time = 2 * time it takes to reach highest point
= 2*(28/9.8)
= 5.71 s
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