Physics 20: Linear Kinematics

Question

A diver springs upward with an initial speed of 1.8 m/s from a 3.0 m board.

a) Find the velocity with which the diver strikes the water. (7.9 m/s down).

b) What is the highest point the diver reaches above the water? (3.2 m).

I want to know how i could find the answer.

Answer 1

To start with you'll have to look at your kinetic equations.

We can use vf2=vi2+2aΔy

Givens:

v_i=1.8 m/s

Δy=3m

g=9.8 m/s^2

a=-g

The way I interpret it you have to find the highest point first. So let's look at part b.

b) We know that at the highest point v_f= 0 m/s, such that:

v_f²=v_i²+2aΔy

Δy=(-v_i²)/(2a)=(-(1.8 m/s)²)/(2(-9.8 m/s²))= .165 m

Add on the total height of the board to be 3m+.165m= 3.165 m ≈ 3.2 m (sig figs)

a)We have all the given quantities. On the way down the speed of the diver is m/s.

We are looking for the final speed reminder that on the way down from the top height velocity is 0 m/s, initial velocity. Also, we have suggested up is positive so going in the down direction Δy is negative as displacement is a vector.

v_f²=v_i²+2aΔy

v_f= ²√(2(-9.8 m/s²)(-3.165 m)) =7.876 m/s ≈ 7.9 m/s down