When the projectile reaches its maximum height its vertical component of velocity will be zero. We can use the linear equations of motion to calculate this height, h.
From where v is the final velocity, u is the initial velocity, a is the acceleration in this case provided by gravity and s is the distance covered or in our case the height h.
The final velocity is zero. So, substituting our symbols we have
But vy is v0 sin θ
Time of Flight
We can use the independence of the horizontal and vertical motions of the projectile to determine the time of flight. If we know the initial velocity, we can work out the time. The vertical motion looks like that of a ball being thrown upward. At its highest point the velocity will be zero. We just need to find the time that it takes and multiply by two, since what goes up, must come down.
Recall from the linear equations of motion, v = u + a t
So, rearranging,
Range
Since the trajectory in the absense of air resistance is symmetrical about the point of maximum height, it will be half way through its motion so the range will be twice the distance it takes to travel in this time.
The horizontal distance is just,
Inserting the horizontal component of velocity and using the argument in the previous section to calculate the time to reach the maximum height, not forgetting to multiply by 2, we obtain The identity can be used to simplify the expression further. If the projectile is fired at some height h0 above the ground that it lands on we must find the additional time that it stays in the air. To find the time involves solving this quadratic equation in t.
We are only interested in the positive solution since a negative value for time is not physical. To check that this expression is correct, it should return to the expression for range on level ground when h0 = 0.
