Question 479546
A ball is thrown vertically upward at a velocity of 15 feet per second from a bridge that is 55 feet above the level of the water. The height h, in feet, of the ball above the water at the time t in seconds after it is thrown is 
h = 16t^2 + 20t - 48
Find the time when the ball strikes the water.
**
When the ball strikes the water, its height above the water=0
h = 16t^2 + 20t - 48=0
solve by quadratic formula as follows:
..
 {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
..
a=16, b=20, c=-48
t=[-20ħsqrt(400-4*16*-48)]2*16
t=[-20ħ√3472]/32
t=(-20ħ58.92)/32
t=1.22 sec
or
t=-2.47 sec (reject,t>0)
ans:
The ball strikes the water 1.22 sec after it is thrown