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Question 95079: If a ball is thrown upward from the roof of a building 80 m tall with an initial velocity of 15 m/s, its approximate height h after t is given by
h = -5t^2 + 15t + 80.
When we used feet, the t^2-coefficient was -16 (from the fact that the acceleration due to gravity is approximately 32 ft/s^2). When we use meters as the height, the t^2-coefficient is -5 (that same accelerations becomes approximately 10 m/s^2). Use this information to solve 31 and 32.
31)Science and Medicine: When will the ball reach a height of 85 m?
32)Science and Medicine: How long does it take the ball to fall back to the ground?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! If a ball is thrown upward from the roof of a building 80 m tall with an initial velocity of 15 m/s, its approximate height h after t is given by
h(t) = -5t^2 + 15t + 80.
When we used feet, the t^2-coefficient was -16 (from the fact that the acceleration due to gravity is approximately 32 ft/s^2). When we use meters as the height, the t^2-coefficient is -5 (that same accelerations becomes approximately 10 m/s^2). Use this information to solve 31 and 32.
31)Science and Medicine: When will the ball reach a height of 85 m?
h(t) = -5t^2 + 15t + 80
85 = -5t^2 + 15t + 80
5t^2-15t+5 = 0
t^2-3t+1 = 0
t = [3+-sqrt(9-4*1*1)]/2
t = [3+-sqrt(5)]/2
Positive answers:
t = [3+sqrt5]/2 = 2.62 seconds (on the way down)
or t =[3-sqrt(5)]/2 = 0.38197 seconds (on the way up)
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32)Science and Medicine: How long does it take the ball to fall back to the ground?
Height is zero when the ball is on the ground
h(t) = -5t^2 + 15t + 80
-5t^2 + 15t + 80 = 0
t^2 -3t-16 = 0
t = [3 +- sqrt(9-4*1*-16)]/2
t = [3+-sqrt73]/2
Positive answer:
t = (3+sqrt73) / 2
t = 5.77 seconds
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Cheers,
Stan H.
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