The first thing to answer part c. Since there is no air resistance then the same acceleration that slows the ball on the way up will also accelerate it on the way down. So, it hits the ground at the same speed of 50.0 m/sec.
Part a is the next simplest. V=at + V_{0}. In this case the initial velocity V_{0} is 50m/sec. At its apex the velocity has become zero (0). The acceleration is gravity a = -9.8m/sec^{2}. Which gives us,
0 = (-9.8m/sec^{2})t + 50m/sec
(9.8m/sec^{2})t = 50m/sec
t = (50/9.8) sec
t = 5.10 sec
Lastly for part b the formula is the standard distance formula: d = (1/2)at^{2} + V_{0}t + d_{0}
Initial distance (d_{0}) is zero (0).
Initial velocity is 50.0 m/s.
Acceleration is gravity -9.8 m/s^{2}.
And the time it took to reach its maximum height is 5.10 sec.
So we get:
d = (1/2)(-9.8 m/s^{2})(5.10s)^{2} + (50.0m/s)(5.10s) + 0
d = (-4.9 m/s^{2})(26.03s^{2}) + (255.10m)
d = (-127.45m) + (255.10m)
d = 127.65 m