SOLUTION: 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 determin
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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.
2)What would be the maximum height of the ball?
3)How long will it take for the ball to hit the ground?
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
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
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
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