A projectile is shot from the ground at an angle of #pi/8 # and a speed of #5 /7 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?
There are two ways to solve this problem. First, the long way.
We first find the vertical and horizontal velocities which are
Next, we must use the equation
to find the time it took for the projectile to reach its maximum height.
We first rearrage to find t, which is
v in this case would be zero, as when the ball reaches it's maximum height, it has 0 vertical velocity. a would be acceleration due to gravity and u would be the value we calculated to initial vertical velocity. Thus we get
This is the time it took for the projectile to reach it's maximum height. We now can find the distance, vertically and horizontally the projectile will travel.
The vertical distance it will travel when reaching it maximum height can be found using
Rearragned, it gives us distance
Substituting the values in to get the verticaldistance, we get
Now to get the horizontal distance, we simply use
whcih gives us
Using pythagoras, we get the total final distance which would be