Answer to Question #129362 in Physics for master

Question

A ball is released at point P with a tangential velocity of 5 ms-1 to move in a circular track in a vertical plane as shown in the Figure 1. Can the ball reach the highest point of the circular track of radius 1.0 m? Give reasons. [4 marks]

Answer 1

$\frac{mv^2}{2}=mgh\to h=\frac{v^2}{2g}=\frac{5^2}{2\cdot9.81}=1.27(m)$

In our case $h=2R=2\cdot1=2(m)$

$2>1.27$

So, the ball can not reach the highest point of the circular track