1. (a) For the first question, are you sure the leap height is correct? 0.112 m is rather weak...
Anyway, you have the leap height and the height of the player, you can get the change in height.
Then use: v^2 + u^2 = 2as
v is the velocity of the player at the peak of the leap (it will be equal to zero at the top), u is the velocity of the player just before leaving the floor, a is the acceleration due to gravity, here equal to -9.81 m/s^2 (or to the appropriate figure your question asks you to use) and s is the change in height.
(b) Use v^2 + u^2 = 2as again, this time, but this time, u will be equal to 0, v will be the velocity you got in (a), s will be a third of the player's height and you have to calculate a to get the acceleration.
Then use this acceleration in either of:
s = ut + 0.5 at^2
v = u + at
to get the time.
2. (a) You will need to generate two equations, one for the car and one fort he police, in terms of distance.
That of the car is obtained from s = ut (constant velocity), the other for the police: s = 0.5at^2
Equate both equations for the distance to be equal to each other after substituting the relevant values to get the time for the police to catch up with the car.
(b) Substitute back this time into any of the two equations you formed in (a).
Post your workings!
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Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet.
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