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# Help, motion questions watch

1. Hi, I think this is meant to be simple, but I have no clue how to do these questions. My teacher zoomed through this topic without explaining it properly.

A cricketer throws a ball vertically upwards so that the ball leaves his hands at a speed of 25ms(-1). If Air resistence can be neglected, calculate:

The maximum height reached by the ball

The time taken to reach maximum height

The speed of the ball when it is at 50% of the maximum height.

For the first question, I have to work out S, and I only know V (and U?) and A (am I meant to use -9.81 here?).I 'm not sure wether I'm meant to use the constant accelaration equations, but I'm stumped.
I would really appreciate a few hints
2. The max height of a body is given by

mgh = 0.5mv^2
gh = 0.5v^2

h = v^2/2g
3. a=-9.8 m/s/s
h=?
v=0 m/s
u= 25 m/s

v^2 = u^2 +2as
0 = 625 -19.6S
-625=-19.6S
s= 31.89m

For part two plug in the values above into s=ut + 1/2at^2

For part 3 divide s (height) from part one by two, and then plug in that value as s, and all the other values apart from v into the original equation.

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Updated: October 14, 2009
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