If a ball is thrown upward at 64 feet per second from the to

If a ball is thrown upward at 64 feet per second from the top of a building that is 100 feet high, the height of the ball is given by

S = 100 + 64t 16t² feet

where t is the number of seconds after the ball is thrown. How long after it is thrown is the height 100 feet?

t=______ s

Solution

S = 100 + 64t - 16t²

substitute S = 100 in the above equation and solve for t. This will give the time taken by the ball to reach the same height as before ( which is 100m above the ground).

100 = 100 + 64t - 16t²

=> 64t - 16t² = 0

=> 16t² - 64t = 0

=> 8t[2t - 8] = 0

which gives: t = 0s and t = 4s.

since t = 0s is the initial condition therefore this value can be discarded.

therefore, t = 4s.