Unsettledchim said:
Thank you and yes that would be great if you could drive the equation
The total velocity can be written as components of the velocity in the x direction and in the y direction:
V(total) = sqrt(V(x)^2 + V(y)^2)
where V(x) = V(0)cos(theta) and V(y) = V(0)sin(theta)
In the x direction, the object travels the following distance (let's call it D(x), which is actually the range R in my last post):
D(x) = V(x)*t = V(0)cos(theta)*t
Our goal is now to find an expression for t, the time, and we can do that by examining the motion in the y direction:
D(y) = V(y)*t - (1/2)*g*t^2 = V(0)sin(theta)*t - (1/2)*g*t^2
Now take the derivative with respect to time and set it equal to zero. This will tell you the time at which the object's velocity in the y direction is zero. Practically, you're finding out when the object reaches the top of the parabolic curve (the greatest height in its motion).
D'(y) = V(0)sin(theta) - g*t = 0
Solve for t:
t = V(0)sin(theta) / g.
The time to complete the entire distance is obviously twice this:
t = 2V(0)sin(theta)/g.
Plug this t back into the D(x) equation to get:
D(x) = 2*V(0)^2*sin(theta)*cos(theta)/g
You can leave the equation in this form if you want, but making the trig substitution sin(2*theta) = 2sin(theta)cos(theta) makes it look nicer:
D(x) = V(0)^2sin(2*theta)/g
