Question 675528
A ball is thrown vertically upward with an initial velocity of 48 feet per second. If the ball started its flight at a height of 8 feet, then its height s at time t can be determined by s(t)=-16t^2+48t+8 where s(t) is measured in feet above the ground and t is the number of seconds of flight.
1)Determine the time it takes for the ball to attain its maximum height.
s(t)=-16t^2+48t+8
complete the square
s(t)=-16(t^2-3t+9/4)+8+36
s(t)=-16(t-3/2)^2+44
ball to attain its maximum height of 44 ft above the ground at 3/2 or 1.5 sec
..
2)What would be the maximum height of the ball?
maximum height of the ball: 44 ft
..
3)How long will it take for the ball to hit the ground?
ball hits the ground when height=0
-16t^2+48t+8=0
use following quadratic formula to solve for t
{{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
a=-16, b=48, c=8
ans:
t≈-0.158 (reject, t>0)
or
t≈3.158
the ball will hit the ground in 3.158 sec