For the 2nd problem, since there is no more data to play with than the givens, I assume the velocity of the ball is due to the motion of the train only. The ball leaves the thrower's hand with a velocity whose horizontal component is the velocity of the train.
Since the we are asked to find how high the ball goes up, then we are interested on the vertical component of the velocity of the ball as the ball leaves the student's hand.
Let v = that vertical component, in m/sec.
And H = maximum height attained by the ball above the tha raised hand of the student, in meters.
t = time for the ball to reach H, in seconds.
tan(50deg) = v/8.5
v = 8.5tan(50deg) = 10.13 m/sec.
H = v*t -(1/2)g*t^2 ------------(i)
To find t, we know that at H, the ball stops moving up, so,
v -gt = 0
t = v/g = 10.13/9.8 = 1.03367 sec.
Substitute that into (i),
H = 10.13*1.03367 -(1/2)(9.8)(1.03367^2) = 5.23556 m.
Therefore, the lady Prof sees the ball rise a maximum height of about 5.24 meters above the palm of the student. --------answer.