A ball is projected at an angle b from the horizontal. With initial velocity V the ball leaves the ground at point O and hits the ground at point A. If Vcos(b) = 6u and Vsin(b) = 2.5u, how long does the ball take to travel between O and A.

This is a classic 2-D projectile question.

The best way to solve this problem is to split the velocity into its x and y components.

In the x-direction a projectile has no forces acting on it (as we neglect air resistance) and so the horizontal velocity remains constant. Remember, a constant velocity means we use the equation V=d/t.

In the y-direction the only force acting is in the negative y direction and is due to constant acceleration of gravity. (F=ma=-mg). Remember, a constant acceleration means we use the SUVAT equations.

First, lets work out the initial velocity components in each direction using the information that we have been given.

Using simple trig we find that:

V(horizontal) = Vcos(b) which we were told is 6u

V(vertical) = Vsin(b) which we were told is 2.5u

Let's look at the vertical component. Remember, a constant acceleration means using the SUVAT equations. But which one? We know the initial (vertical) velocity, the (vertical) displacement must be 0 as the projectile starts and ends on the ground, we know the (vertical) acceleration is –g and we want to find the time taken.

So, we need the SUVAT equation that includes u, s, a and t:

s = ut + (1/2)at^2

Now lets substitute in our values and rearrange:

0 = 2.5ut -(1/2)gt^2

Cancel a t:

0 = 2.5u – (1/2)gt

Finally, rearrange for t:

t = 5u/g