Abaseball player throws a 95mph fastball straight up into the air. the position equation, which gives the height of the ball at any time t, in seconds, is given by: s(t)=-16t^2+140t+37 a.) find the maximum height of the ball. b.) find the time and velocity when the ball hits the ground.

i would really appreciate it if someone explained to me how to solve this!

Set s(t) = 0 which means it hits the ground.  The formula doesn't fit the parameters given as it shows that the pitcher is standing on something 37 feet high.
s(t)=-16t^2+140t+37 and has an initial velocity of 140.
Graphing or solving this, t= 9 seconds when it hits the ground.
The velocity V(t) is the derivative of s(t)
V(t) = -32t+140
V(9) = -148 ft/second which is going down