Relative Velocity Homework: Solve for Ball Rise Height

[artfuldodger2]

Homework Statement

A science student is riding on a flatcar of a train traveling along a straight horizontal tract at a constant speed of 10.0m/s. The student throws a ball along a path that she judges to make an initial angle of 60.0degrees with the horizontal and to be in line with the track. The student's professor, who is standing on the ground nearby, observes the ball to rise vertically. How high does the ball rise?

Homework Equations

velocity(ab) = velocity(ae)-velocity(be)
change in y=(v{o} sin{theta})t-.5gt^2
= v{o}t+.5at^2
change in x=v{o}t

The Attempt at a Solution

I'm not sure what to assign for ab, ae and be in this problem.
Also, once I solve for the actual velocity, I'm still left with two unknowns, how far it has gone and how long it took to get there. What am I missing here?

 

[Kurdt]

Since the professor just observes the ball to rise vertically one can assume that the horizontal component of the balls velocity is 10 m/s in the opposite direction to the trains motion.

 

[artfuldodger2]

Thank you for your reply Kurdt.

I was able to figure out the problem this morning after a good nights rest :)


I found the velocity in the vertical direction and from there it's pretty simple to find the max height using

h= v{y}^2(sin^2(theta)) / 2g

and in this case theta was 90, so it was just the velocity in the y-direction squared, over 2g.


At least, I think that's how it's done..?

 

[Kurdt]

Yeah that looks good.