Question

A person throws a ball straight up. He releases the ball at a height of 1.75 m above the ground and with a velocity of 12.0 m/s. Ignore the effects of air resistance. (a) How long until the ball reaches its highest point? (b) How high above the ground does the ball go?

Answer 1

a) 1.22 s

b) 9.089 m

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

`v=u+at\\\Rightarrow t=(v-u)/(a)\\\Rightarrow t=(0-12)/(-9.81)\\\Rightarrow t=1.22\ s`

Time taken by the ball to reach the highest point is 1.22 seconds

`s=ut+(1)/(2)at^2\\\Rightarrow s=12* 1.22+(1)/(2)* -9.81* 1.22^2\\\Rightarrow s=7.339\ m`

The maximum height the ball will reach above the ground is 1.75+7.339 = 9.089 m