A ball is thrown upward with an initial velocity of 64 ftsec

A ball is thrown upward with an initial velocity of 64 ft/sec from a height of 960 ft. It\'s height h, in feet, after t seconds is given by h(t)= -16t^2+64t + 960. After how long will the ball reach the ground?
The ball will reach the ground in __ seconds.

Solution

When the ball strikes the ground, h(t) = 0, therefore
-16t^2 + 64t + 960 = 0
;
Simplify & change the signs, divide thru by -16:
t^2 - 4t - 60 = 0
:
Factor
(t - 10)(t + 6) = 0
:
Positive solution:
t = 10 sec; The ball will reach the ground
;
:
Check solution in original equation
h(t) = -16(10^2) + 64(10) + 960
h(t) = -1600 + 640 + 960
h(t) = 0; confirms our solution