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mechanics question - will give rep.

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

Reply 1

Nylex

10

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.

Reply 2

elpaw

15

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

Reply 3

ShOcKzZ

OP

7

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 :biggrin:

i'll rep you tmrw
thanks for helping.

Reply 4

elpaw

15

ShOcKzZ

correct answer but i don't understand any of it :biggrin:

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)

Reply 5

ShOcKzZ

OP

7

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.

Reply 6

elpaw

15

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

Reply 7

Nylex

10

Lol, thanx for rep (both of you) :smile:

.

mechanics question - will give rep.

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