Question

A ball with a mass of `480 g` is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of `32 (kg)/s^2` and was compressed by `7/8 m` when the ball was released. How high will the ball go?

Answer 1

The height is `=2.6m`

Explanation:

The spring constant is `k=32kgs^-2`

The compression is `x=7/8m`

The potential energy is

`PE=1/2*32*(7/8)^2=49/4=12.25J`

This potential energy will be converted to kinetic energy when the spring is released

`KE=1/2m u^2`

The initial velocity is `=u`

`u^2=2/m*KE=2/m*PE`

`u^2=2/0.48*12.25=51.04`

`u=sqrt51.04=7.14ms^-1`

Resolving in the vertical direction `uarr^+`

We apply the equation of motion

`v^2=u^2+2ah`

At the greatest height, `v=0`

and `a=-g`

So,

`0=51.04-2*9.8*h`

`h=51.04/(2*9.8)=2.6m`