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# mechanics question - will give rep. watch

1. can't do this dam question although im sure its not too difficult.

tennis ball hit so that it moves vertically downwards from a height of 1m with an initial speed of 5ms^-1. when it hits the ground it rebounds vertically with half the speed it had when it hit the ground.

Q - find the height to which it rebounds
2. (Original post by ShOcKzZ)
can't do this dam question although im sure its not too difficult.

tennis ball hit so that it moves vertically downwards from a height of 1m with an initial speed of 5ms^-1. when it hits the ground it rebounds vertically with half the speed it had when it hit the ground.

Q - find the height to which it rebounds
v^2 = u^2 + 2as

v = (u^2 + 2gs)^1/2

v = (25 + 2 x 9.8 x 1)^1/2

Half that answer then use that as your initial speed and v = 0.

v^2 = u^2 + 2as

=> s = (v^2 - u^2)/2g

Remember to use g = -9.81.
3. Just using conservation of energy (taking g=9.81):

mg1+m5²/2 = mv²/2 => v = 6.68 m/s when it hits the ground

m(3.34)²/2 = mgh => h = 0.57m
4. (Original post by elpaw)
Just using conservation of energy (taking g=9.81):

mg1+m5²/2 = mv²/2 => v = 6.68 m/s when it hits the ground

m(3.34)²/2 = mgh => h = 0.57m
correct answer but i don't understand any of it
i'll rep you tmrw
thanks for helping.
5. (Original post by ShOcKzZ)
correct answer but i don't understand any of it
i'll rep you tmrw
thanks for helping.
dont rep me, rep nylex, he's got a correct method too. i believe my method is M3, so you wont get it just yet (if you havent done M3)
6. (Original post by elpaw)
dont rep me, rep nylex, he's got a correct method too. i believe my method is M3, so you wont get it just yet.
i did rep nylex
but i will rep you for takin the time to reply..
and im doing M3, but i its AQA B so its like M1 really..if that makes sense.
7. (Original post by ShOcKzZ)
i did rep nylex
but i will rep you for takin the time to reply..
and im doing M3, but i its AQA B so its like M1 really..if that makes sense.
oh ok, well its just conservation of energy, where:

(GPE + KE)start = (GPE + KE)end
where GPE = mgh and KE = mv²/2

so for the first bit:

mg1 + m5²/2 = mg0 + mv²/2
m's cancel, so you get v²/2 = 9.81 + 12.5 = 22.36 => v²=44.72 => v = 6.68 m/s

its speed is halved upon impact, so v = 3.34 m/s

using CoE again:

mg0 + m(3.34)²/2 = mgh + m0²/2
m's cancel, so you get 5.5775 = 9.81h => h = 5.5775/9.81 = 0.57 m
8. Lol, thanx for rep (both of you) .

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