SOLUTION: at 9:45am Margie threw a ball upwards while standing on a platform 35 ft. above the ground.
The height after t seconds follows the equation:h(t)= -0.6t^2+72t+35
a.) What will the
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-> SOLUTION: at 9:45am Margie threw a ball upwards while standing on a platform 35 ft. above the ground.
The height after t seconds follows the equation:h(t)= -0.6t^2+72t+35
a.) What will the
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Question 572577: at 9:45am Margie threw a ball upwards while standing on a platform 35 ft. above the ground.
The height after t seconds follows the equation:h(t)= -0.6t^2+72t+35
a.) What will the maximum height of the ball be?
b.)How long will it take the ball to reach its maximum height?
I have tried this problem a few times and keep getting 107.36ft. I dont think im doing the set up correctly. Thanks for the help! Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! at 9:45am Margie threw a ball upwards while standing on a platform 35 ft. above the ground.
The height after t seconds follows the equation:h(t)= -0.6t^2+72t+35
a.) What will the maximum height of the ball be?
b.)How long will it take the ball to reach its maximum height?
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b.)How long will it take the ball to reach its maximum height?
h(t)= -0.6t^2+72t+35
Max height is the vertex of the parabola.
It's on the LOS (Line of Symmetry) t = -b/2a
t = -72/(-1.2)
t = 60 seconds
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a.) What will the maximum height of the ball be?
Max ht is h(60)
h(60) = -0.6*60^2 + 72*60 + 35
= 2195 feet
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The equation gives a gravity constant of 1.2 ft/sec/sec.
On Earth, it's 32 ft/sec/sec . I wonder where Margie lives.
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The 945 AM is not relevant.