A cannon ball is fired directly upward with a velocity of 215 m/s. How long does it take to fall back to the ground? s How fast is it going when it hits the ground? m/s

Question

a. A cannon ball is fired directly upward with a velocity of 215 m/s. How long does it take to fall back to the ground?   s

b. How fast is it going when it hits the ground?   m/s

Answer 1

If v = velocity (assuming no air resistance0

v_o = initial velocity (215 m/s)

t = time

g = acceleration due to gravity = 9.8 m/s²

then,v = v_o - gt and at the max height v = 0, then t = 21.94 seconds up so

t = 43.87 seconds to fall back to the ground

Looking at the trip up, y = v_ot - (1/2)gt² or y = 2358 m

The velocity hitting the ground can be calculated using the energy formula v = sqrt(2*g*h) where h is y (the height) in this case

.

So, v = sqrt(2*9.8*2358) = 215 m/s which is the same as the initial velocity!