Question

Someone is jumping from a 1m high stage. The slow down took place after 0,5s. How high is the uniform negative acceleration?

Answer 1

Why is the equation \[ a= v/t = \left( 2sg \right)^{1/2} / t = (2*1m*9,81ms ^{-2})^{1/2} /3,6*5s\] ?

Answer 2

So where does the 1/2 come from and why is it 3,6*5s and not 3,6*0,5s ?

Answer 3

What do you mean when you say "The slow down took place after 0,5s" ??

Answer 4

The braking operation.