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

Answer:

5.37 m/s

0.98 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -9.81\times 10+15^2}\\\Rightarrow v=5.37\ m/s$

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

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{5.37-15}{-9.81}\\\Rightarrow t=0.98\ s$

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