Question

A cannon ball is launched at 6 feet per second from the top of a platform 12 feet above the ground.

Answer 1

A.) y = ut - 1/2gt^2 + 12
y = 6t - 1/2(9.8)t^2 + 12
y = 6t - 4.9t^2 + 12

b.) The y-intercept is 12

c.) The y-intercept represent the height of the ball at time t = 0

d.) At the time the ball reaches the ground, y = 0
-4.9t^2 + 6t + 12 = 0
t = 2.29 seconds
Therefore, it took the ball 2.29 seconds to reach the ground.