Question

vo 5 m/s A ball is hanging straight down from a string with a length of 2 m. If the ball is pushed so that it has an initial velocity of 5 m/s, how high above the ball's lowest point will the ball reach?

Answer 1

Concept used :

Mechanical energy conservation

Solution:

After getting pushed the only force that is doing work on ball is the gravitational force so we can say that mechanical energy of the ball will remain conserved.

Initial mechanical energy = kinetic energy = mv²/2

Mechanical energy at highest point = Gravitational potential energy = mgh .......(h = maximum height from initial point)

So from mechanical energy conservation we can say that

$mgh = \frac{mv^2}{2}$

$h = \frac{v^2}{2g}$

Given: v = 5 m/s , g = 9.8 m/s²

$h = \frac{(5m/s)^2}{2\times 9.8 m/s^2}$

h = 1.275 m .................Answer