Question

A diver springs upward with an initial speed of 1.75 m/s from a 3.0-m board. (a) Find the velocity with which he strikes the water. [Hint: When the diver reaches the water, his displacement is y = -3.0 m (measured from the board), assuming that the downward direction is chosen as the negative direction.] (b) What is the highest point he reaches above the water?

Answer 1

a) 7.848 m/s

b) 3.156 m

Step-by-step explanation:

t = Time taken

u = Initial velocity = 1.75 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

`v=u+at\\\Rightarrow 0=1.75-9.81* t\\\Rightarrow (-1.75)/(-9.81)=t\\\Rightarrow t=0.18\ s`

`s=ut+(1)/(2)at^2\\\Rightarrow s=1.75* 0.18+(1)/(2)* -9.81* 0.18^2\\\Rightarrow s=0.156\ m`

His highest height above the board is 0.156 m

b) Total height he would fall is 3+0.156 = 3.156 m

`s=ut+(1)/(2)at^2\\\Rightarrow 3.156=0t+(1)/(2)* 9.81* t^2\\\Rightarrow t=\sqrt{(3.156* 2)/(9.81)}\\\Rightarrow t=0.8\ s`

He is in the air for 0.8+0.18 = 0.98 seconds

`v=u+at\\\Rightarrow v=0+9.81* 0.8\\\Rightarrow v=7.848\ m/s`

a) His velocity when he hits the water is 7.848 m/s