is anyone kind enough to help me w/ this physics?

Question

a bullet is fired straight up with an initial velocity of 250 meters per second. (a) by setting the motion equation equal to zero, find how long it takes for the bullet to have a vertical velocity of zero. (b) using the time found from part (a), find the maximum height the bullet reaches.

Answer 1

h = ut - 1/2 gt^2; where h = height, u = initial velocity = 250 m/sec, g = 9.81 m/sec^2, and t = time to rise to h.

v = u - gt; so that, when v = 0, velocity at h, u = gt and t = u/g = 250 m/sec / (9.81 m/sec^2).

Substitute u/g into first SUVAT equation to get h = u^2/g - 1/2 g(u/g)^2 = u^2/g - 1/2 (u^2/g) = 1/2 (u^2/g) = 1/2 (250^2/9.81) (You can do the math.)

h in this case is measured from the gun muzzle, not the Earth's surface. To find the height above Earth's surface you need to add in the height of the shooter's hand when the gun was fired.