Question

A ball is batted straight up into the air and reaches a maxium height 65.6 m (a) How long did it take to reach this height? (b) What was the pop-up velocity of the ball?

Answer 1

a) 3.65 seconds

b) 35.87 m/s

Step-by-step explanation:

s = Displacement = 65.6 m

u = Initial velocity

v = Final velocity

t = Time taken

a = Acceleration due to gravity = 9.81 m/s² (downward direction is taken as positive and upward is taken as negative)

b) Equation of motion

`v^2-u^2=2as\\\Rightarrow 0^2-u^2=2* -9.81* 65.6\\\Rightarrow u=√(2* 9.81* 65.6)\\\Rightarrow u=35.87\ m/s`

Initial pop up velocity is 35.87 m/s

a)

`v=u+at\\\Rightarrow t=(v-u)/(a)\\\Rightarrow t=(0-35.87)/(-9.81)\\\Rightarrow t=3.65\ s`

It took 3.65 seconds to reach this height