A person is riding on a flatcar traveling at a constant speed v1= 20 m/s with respect to the ground. He wishes to throw a ball through a stationary hoop in such a manner that the ball will move horizontally as it passes through the hoop. The hoop is at a height h=4 m above his hand. He throws the ball with a speed v2= 18 m/s with respect to the flatcar. Let g=10 m/s2 and neglect air drag completely. (see figure)
(a) At what horizontal distance x in front of the hoop must the person release the ball? (in meters)
(b) When the ball leaves his hand, what is the direction of the velocity vector of the ball as seen from the flatcar? (angle αcar with respect to the horizontal in degrees)
(c) When the ball leaves his hand, what is the direction of the velocity vector of the ball as seen from the ground? (angle αground with respect to the horizontal in degrees
The Attempt at a Solution
I found the y component of the initial velocity using the equation 0=v_0*2=2(-10)(4), which I found to be 4sqrt(5). I used this to find the x component of the initial velocity which I calculated to be 2sqrt(61). I found the angle with respect to the boxcar by taking sin^-1=4sqrt(5)/18), angle =29.795 degrees.
To find the time to reach the hoop I used the equation 4=4sqrt(5)-5t^2, t=.8944. Knowing the time and x-component of the velocity I calculated the horizontal distance to be 13.971.
I have no clue how to find the angle relative to the boxcar.
Can someone please check my work on the first 2 parts and how to get the third part? Thanks!