Answer to Question #296305 in Physics for Kazi

Question

A cliff diver from the top of a 100 m cliff. He begins his dive by jumping up with a velocity of 5 m/s. What is his velocity right before he hits the water?

a. 22.3 m/s

b. 11.15 m/s

c. 1,989 m/s

d. 44.6 m/s

 

Answer 1

Given:

$v_1=5\:\rm m/s$

$h=\rm 100\: m$

The law of conservation of energy says

$\frac{mv_1^2}{2}+mgh=\frac{mv_2^2}{2}$

Thus, the final velocity

$v_2=\sqrt{v_1^2+2gh}=\sqrt{5^2+2*9.8*100}=\rm 44.6\: m/s$

Answer: d. 44.6 m/s