Question

A particle is thrown vertically upwards with an initial velocity . Find the [3]
maximum height h , to which the particle rises .

Answer 1

Answer:

To answer this question we have to make a few assumptions. I’m going to model the ball as a particle with no air resistance or external contributions to the energy of the ‘ball’. I will also take g (the acceleration due to gravity) as  9.81ms−2 .

Explanation:

There are multiple ways of doing this, I will start by considering energy and then using SUVAT.

Energy

The ball is given kinetic energy when it is launched which, under my assumptions, is all converted to gravitational potential energy and the ball is at its maximum height when it runs out of kinetic energy and it has all become gravitational potential energy.

We can therefore say that kinetic energy  12mv2  is equal to gravitational potential energy,  mgh  

12mv2=mgh  

12v2=gh  

12v2g=h  

v22g=h=25.522×9.81≈33.14m  

SUVAT

We know the ball will be traveling at  0ms−1  when at its highest point because the velocity will decrease on its way up and then reach 0 where it will decrease into negative velocity on its way down, using up as the positive direction.

For those of you not aware of the SUVAT equations, S represents the displacement, U and V are initial and final velocities respectively, A is acceleration and T is time. These are all properties of an object in motion. There are five equations which using any 3 variables we can find the other 2 variables using the equations as long as the acceleration of the object is constant.

We have U -  25.5ms−1 , V -  0ms−1  and A -  −9.81ms−2  and we want to find S, we are not interested in T.

We can use the equation

V2−U2=2AS  and rearrange to find S

V2−U22A=S=02−25.522×−9.81≈33.14m  

As you can see both methods returned the same result.