Asked by Tara psd
A ball is thrown from a height of 10m with certain initial downward velocity. It hits the ground and loses 50% of its energy during the impact and rises back to the sane height. What was its initial downward velocity of projection?
Answers
Answered by
Anonymous
initial energy
= (1/2)m v^2 + m g (10)
= m(.5 v^2+98.1)
final energy = m(98.1)
so
98.1 = .5(.5v^2 +98.1)
2(98.1) = .5 v^2 + 98.1
.5 v^2 = 98.1
v = sqrt(2*98.1)
v = about 14 m/s
= (1/2)m v^2 + m g (10)
= m(.5 v^2+98.1)
final energy = m(98.1)
so
98.1 = .5(.5v^2 +98.1)
2(98.1) = .5 v^2 + 98.1
.5 v^2 = 98.1
v = sqrt(2*98.1)
v = about 14 m/s
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.