# Maximum Height

You've already found out about how to find the time of flight of the projectile, and now you're about to learn how to find the projectile's maximum height. The maximum height of a projectile's flight can be calculated using the formula v2 = u2 + 2as and is again entirely reliant on the initial vertical velocity, as gravity would serve to apply a negative acceleration on the projectile. As before, v = 0 (that is, final velocity is zero at the maximum point of projection), u = uv, a = -9.8 m/s2

That is, 0 = uv2 - 19.6s
uv2 = 19.6s
s = uv2 / 19.6

 s = uv2 / 19.6   or  s = (usinx)2 / 19.6 where,  s = displacement (m)  uv = initial vertical velocity (m/s)  u = initial velocity of object (m/s)  x = angle of projection (degrees)

## Examples

Q: Tang is launched with a velocity of 50m/s at 45 degrees to the ground. Assuming gravity is 9.8 m/s2, and we already found out last time his total time of flight, can you figure out how high he'll get?

A:

To find the answer you should utilise the formula v2 = u2 + 2as.

At the highest point of his flight Tang would have a vertical velocity of 0m/s while gravity exerts an acceleration of -9.8m/s2 on his body as well. His initial vertical velocity is already found out to be 50 x sin 45 m/s.
Therefore for our equation we can put v = 0m/s, and a = -9.8m/s2, u = 50 x sin 45.

Substituting the values in we get:

0 = 1250 - 2 x 9.8 x s
19.6s = 1250
s = 63.776 metres (3 decimal places)

Therefore the maximum height that Tang reaches in this flight is approximately 64 metres.

Congratulations!! You have just completed your second lesson in Projectile Motion. You are now an expert on finding the maximum height flight of a projectile. Its time for you to show us how much you have learnt from this lesson. Brace yourself for these 5 questions. And if can do them you can go to the next lesson - finding the horizontal range of flight.

### Questions

1) At the point of maximum height, a projected object is:

a) stationary.
b) moving with a positive velocity.
c) moving with a negative velocity.
d) None of the above.

2) Which of the following formulae are useful in determining maximum height?

a)v = u + at.
b) F = ma.
c) v x v = u x u + 2as.
d) PE = mgh.

3)What is the maximum height reached by an object which is launched vertically upwards at 25m/s?

4) A pineapple is fired with an initial velocity of 50m/s at an angle of 35 degrees above the horizon. What is the maximum height reached?

5) A stone is thrown vertically upward with an inital velcoity of 62.5m/s. Calculate the it's maximum height.