Question

A ball it hit straight up in the air with an initial velocity of 12.6 m/s. At what time does it cross the height of 1 m going up?

Answer 1

Answer:

t=0.08 seconds or 2.49 seconds

Explanation:

using x= ut + 0.5at^2

x = 1

u=12.6m/s

a= -9.8m/s^2. (since it's moving upward )

1=12.6t - (0.5×9.8× t^2)

1= 12.6t - 4.9 t^2

4.9 t^2-12.6t+1=0

solving for t using quadratic formula

t=0.08 seconds or 2.49 seconds