A baseball is hit straight up and is caught by the catcher 2.0 s later. The maximum height of
the ball during this interval is:

Question

A baseball is hit straight up and is caught by the catcher 2.0 s later. The maximum height of
the ball during this interval is:

raffle_snaffle · Answer

There isn't anymore information given?

anonymous · Answer

no

ineedserioushelp · Answer

The answer is: 12.6m . I had this question before and to find the answer use this: d = (1/2)at^2 with t = 1 second and a = 9.8 m/sec^2 . It took 1 second to go in the air at its top speed and 1 second to fall down.

anonymous · Answer

the answer is 4.9

anonymous · Answer

i tried to use max height equation 1/2 V^2(initial in respect to y)/ g

ganeshie8 · Answer

4.9 is correct !

anonymous · Answer

how did you get it????

ganeshie8 · Answer

easy,
since it took 2s for going up and falling down,
it took 1s for going up

ganeshie8 · Answer

agree so far ?

anonymous · Answer

yes!

ganeshie8 · Answer

Next, when it reaches top most, its velocity becomes 0
so,

v = u + at

0 = u -g(1)

u = g = 9.8 m/s

ganeshie8 · Answer

so initial velocity  = u = 9.8 m/s

anonymous · Answer

wait how is acceleration negative?

ganeshie8 · Answer

now, use the position equation :-

S  = ut + 1/2 at^2
 
S = 9.8(1) - 1/2 g (1)^2

S = 9.8 - 1/2 *9.8

S = 9.8 - 4.9

ganeshie8 · Answer

very good q

anonymous · Answer

how i'm setting things up is up is positive and down is negative

ganeshie8 · Answer

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ganeshie8 · Answer

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