Falling objects rate of acceleration

Question

A diver hits the water at a velocity of 17.64 m/s from a 16 m high board. What is the diver's rate of deceleration when the diver reaches a depth of 1.85 m under water?

Answer 1

Hey there Gnarls,

This question is phrased as if the diver hasn't fully decelerated by the time he reaches that depth, but we can only do this problem if we assume that the diver has slowed to a stop when they reach 1.85 m. 

If we do, we can use this equation

 

v_f² = v_i² + 2aΔx

 

v_f = 0m/s

v_i = 17.64m/s

Δx = 1.85m

a = ?

 

Plug it all in we get

311.1696 m²/s² = 0 + 2*a*(1.85 m) Remove 0

311.1696 m²/s² =2*a*(1.85 m) Divide by 2

155.5848 m²/s² =a*(1.85 m) Divide by 1.85 m

84.09989 m/s² = a Round appropriately

 

So there you have it

 

The diver's acceleration was 84.10 m/s²