SOLUTION: When a ball is kicked in the air, its height in meter above the ground can be modeled by h(t)= -4.9t^2 + 14.7t and the distance it travels can be modeled by d(t)=16t where t is the
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Question 695141: When a ball is kicked in the air, its height in meter above the ground can be modeled by h(t)= -4.9t^2 + 14.7t and the distance it travels can be modeled by d(t)=16t where t is the time in seconds.
a. How long is the ball in the air
b. How far does it travel before it hits the ground
c. What is the maximum height of the ball
Can you please help?
Thank you!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
When a ball is kicked in the air, its height in meter above the ground can be modeled by h(t)= -4.9t^2 + 14.7t and the distance it travels can be modeled by d(t)=16t where t is the time in seconds.
a. How long is the ball in the air
The ball starts on the ground and ends up on the ground.
It's height on the ground is zero.
Solve: -4.9t^2+14.7t = 0
t(-4.9t+14.7) = 0
t = 0 and t = 14.7/4.9 = 3 seconds
Ans: The ball in in the air for 3 seconds
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b. How far does it travel before it hits the ground
Ans: 3 meters
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c. What is the maximum height of the ball
Vertex occurs at t = -b/(2a) = -14.7/(2*-4.9) = 3/2
h(3/2) = -4.9(3/2)^2 + 14.7(3/2) = 11.025 meters (maximum height)
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Cheers,
Stan H.
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