A projectile is shot from the ground at an angle of #pi/4 # and a speed of #5 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?
Find the time it takes to reach the maximum height, and use the x component of the displacement equation to find the distance travelled in that time.
First we break up the velocity into it's vector components at time 0,
Since we're presumably ignoring air resistance, the acceleration components are
So our functions for
We can find the time it takes to reach the maximum height by setting
Then we use the displacement formula
and substituting for t yields:
This gives the range of the projectile at the maximum height.
It's interesting to note that this is exactly half the range equation for a projectile returning to the same height.
which is derived using
which is the displacement equation for y.