A golfer stands 390 ft (130 yards) horizontally from the hole and 40 feet below the hole. Assuming the ball is hit with an initial speed of 150 ft/s, at what angle should it be hit to land in the hole? Assume the path of the ball lies in a plane
Force of gravity:
a(t) = <0, -32>
32 feet/sec^2 will be the force of gravity)
Integrating a(t) gives us v(t), the velocity function.
V(t) = <0, -32t> + <Initial Velocity in X Direction, Initial Velocity in Y Direction>
The Attempt at a Solution
1. I obtain the vector valued function for the velocity of the ball. Let's call that v(t). We know that after the ball is launched, there are no forces acting on the ball in the x-direction, but there is gravity acting on the ball in the y-direction. Thus, the y-component of the vector will be -32t, and the x-component will be 0. The initial velocity vector will be the x and y-components of the initial speed of 150 ft/sec, which will be 150cos(θ), and 150sin(θ), respectively.
v(t) = <0, -32t> + <150cos(θ), 150sin(θ)>
v(t) = <150cos(θ), 150sin(θ)-32t>
2. I obtain the vector valued function for the position of the ball by integrating the velocity function. Let's call the position function p(t):
p(t) = <150sin(θ), -150cos(θ)-16t^2> + <Initial Position on X-Coord, Initial position on Y-Coord>
Let's assume the golfer is at the origin, so that the initial position vector is <0,0>.
Thus, p(t) = <150sin(θ), -150cos(θ)-16t^2> + <0, 0> = <150sin(θ), -150cos(θ)-16t^2>
3. Now, do we set the position function equal to the vector <390, 40>, and somehow solve for theta? Any advice on how exactly I can do that?