Question

A ball is hurled straight up at a speed of 15 m/s, leaving the hand of the thrower 2.00 m above the ground. Compute the times and the ball’s speeds when it passes an observer sitting at a window in line with the throw 10.0 m above the point of release.

Answer 1

5.37 m/s

0.98 seconds

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^2-u^2=2as\\\Rightarrow v=√(2as+u^2)\\\Rightarrow v=√(2* -9.81* 10+15^2)\\\Rightarrow v=5.37\ m/s`

Velocity of the ball when it passes an observer sitting at a window is 5.37 m/s

`v=u+at\\\Rightarrow t=(v-u)/(a)\\\Rightarrow t=(5.37-15)/(-9.81)\\\Rightarrow t=0.98\ s`

Time taken by the ball to pass the observer sitting at a window is 0.98 seconds