SOLUTION: A ball thrown in the air with a velocity of 24.6 m/s is at a height represented by h(t)=24.6t-4.9t^2 + 2 meters after t seconds.
a) After how many seconds will the ball fall to th
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Question 254699: A ball thrown in the air with a velocity of 24.6 m/s is at a height represented by h(t)=24.6t-4.9t^2 + 2 meters after t seconds.
a) After how many seconds will the ball fall to the ground? (I think 2)
b) When is the ball increasing? Decreasing?
c) After how many seconds is the height of the ball the greatest?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A ball thrown in the air with a velocity of 24.6 m/s is at a height represented by h(t)=24.6t-4.9t^2 + 2 meters after t seconds.
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Note: h(t) is the height of the ball after t seconds.
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a) After how many seconds will the ball fall to the ground?
The height is zero when the ball is on the ground.
Solve -4.9t^2+24.6t+2 = 0
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I graphed it to get:
t = 5.10 seconds
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b) When is the ball increasing? Decreasing?
The vertex is at t = -b/2a = -24.6/(2(-4.9)) = 2.51 seconds
Increasing: 0<= t <= 2.51 seconds
Decreasing: 2.51< t < 5.10 seconds
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c) After how many seconds is the height of the ball the greatest?
That's what the vertex tell you: t = 2.51 seconds
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Cheers,
Stan H.
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