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EconLou
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#1
Report Thread starter 14 years ago
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Ok, this was a question on my prep last night, and even though i have handed it in (and probs got it wrong!) it is going to irritate me unless someone explains how it should be done!

A girl throws a ball vertically upwards with speed 8m/s from a window which is 6m above horizontal ground.
1.5 secs later she drops a second ball from rest out of the same window
Find the distance below the window of the point where the two balls meet.


Thanks!
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dvs
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Report 14 years ago
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For the first ball, find its velocity at t=1.5. This is the instant when the second ball is dropped. Now, find the displacement of both balls using s=ut+0.5at² (remember to use the new initial velocity for ball 1), and find the time when s of ball 1 = s of ball 2. This is the time when they meet. Use this value of t to find the displacement of the first ball from the window now using the initial velocity given in the question.

I usually avoid doing mechanics questions online, because I prefer drawing diagrams and stuff. So if there's an error in my method, it's not my fault.
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Aitch
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(Original post by dvs)
For the first ball, find its velocity at t=1.5. This is the instant when the second ball is dropped. Now, find the displacement of both balls using s=ut+0.5at² (remember to use the new initial velocity for ball 1), and find the time when s of ball 1 = s of ball 2. This is the time when they meet. Use this value of t to find the displacement of the first ball from the window now using the initial velocity given in the question.

I usually avoid doing mechanics questions online, because I prefer drawing diagrams and stuff. So if there's an error in my method, it's not my fault.
Yes. I get (after 1.5s) ball1 is at +.975m at a v of -6.7 (i.e. down)

forming 2 equations gives

S1 + 0.975 = 6.7t + 4.9 t^2
S2= 4.9 t^2

S1=S2

subtract the second from the first to get 0.975 = 6.7t

t=0.1455
put this value back into (2) to get (I think) 0.104m

Aitch
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