posted by Matthew on .
The height above the ground of a ball thrown up with a velocity of 96 feet per second from a height of 6 feet is 6+96t-16^2 feet, where t is the time in seconds. According to this model, how high is the ball after 7 seconds? Explain (Thank you in advance! :)
6 = 6 + 96t - 16t^2
16t^2 - 96t = 0
16t(t - 6t)=0
t = 0 or t = 6
the ball was 6 feet high at the beginning (t=0) and
on it way back down at 6 seconds later.
(I bet the ball was at its highest point after 3 seconds)