Physics Problem Help- Basbeall trajectory?

Question

During a babseball game a batter hits a pop up to a fielder 72m away. The accleration of gravity is 9.8m/s2. If the ball reamins in the air for 6.2s, how high does it rise.

Answer 1

As you didn't give it, I'm assuming no air drag. Then vx = constant = S/t = 72 m/6.2 sec. Thus, the total energy at h, the max height above ground = TE(h) = KE(x) + PE(h) = 1/2 mvx^2 + mgh. vy, the Y direction velocity = 0 at h max. And the total energy just before the outfielder digs the ball out of the turf = TE(h = 0) = 1/2 mv^2 = 1/2 m(vx^2 + vy^2).

From the conservation of energy we have TE(h) = 1/2 mvx^2 + mgh = 1/2mvx^2 + 1/2mvy^2 = TE(h = 0); so that mgh = 1/2 mvy^2 and h = vy^2/2g. So we need vy, the initial Y velocity off the bat. That's where all the kinetic energy comes from.

We can get vy from uy = vy - gt/2; where uy = 0 at max h; so that vy = gt/2 and h = (gt/2)^2/2g = g^2 t^2/4//2g = 1/8 gt^2 ~ 10*36/8 ~ 50 meters (you can do the math for more precise numbers). We assume t/2, where t = 6.2 sec because that's half the time going up and half the time coming back down.