(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Gravity is assumed to be 10ms^{-2}throughout.

A ball is thrown vertically upward with a speed of 14ms^{-1}. Two seconds later, a ball is dropped from the same point find where the two balls meet.

(They will meet when they are the same displacement (s) from the starting point. If the time taken for ball 1 is T, then the time taken for the second ball is T - 2 seconds.)

2. Relevant equations

v=u+at

v^{2}=u^{2}+2as

s=ut+(1/2)at^{2}

s=(1/2)(u+v)t

3. The attempt at a solution

The answer is [itex]8.\dot{8}[/itex] and the teacher showed another method for solving it but I felt I was getting somewhere with my method but I couldn't figure out what to do next.

At t=0 the variables for ball 1 are:

s=0m

u=14ms^{-1}

a=-10ms^{-2}as the force of gravity is opposing motion.

So then I can work out the velocity and displacement of ball one at t=2.

v=u+at

v=14-10*2

v=-6ms^{-1}so its moving downwards.

s=(1/2)(u+v)*t

s=(1/2)(14-6)*2

s=(1/2)(8)*2

s=8m above the starting position.

This is really as far as I got any meaningful answers as I couldn't figure out what to do next so I just copied down the example method from the black board.

Thanks

Al

PS. I can do a couple of diagrams if anyone needs them to visualise the problem.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Vertical motion under gravity - SUVAT equations.

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