SOLUTION: A ball is thrown straight upward with an initial velocity of 52 ft/sec. Its height above the ground after seconds is given by the formula h(t)=-16t^2+52t. Whta is the maximum heigh
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-> SOLUTION: A ball is thrown straight upward with an initial velocity of 52 ft/sec. Its height above the ground after seconds is given by the formula h(t)=-16t^2+52t. Whta is the maximum heigh
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Question 595628: A ball is thrown straight upward with an initial velocity of 52 ft/sec. Its height above the ground after seconds is given by the formula h(t)=-16t^2+52t. Whta is the maximum height that the ball attains before hitting the ground (feet)? Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! A ball is thrown straight upward with an initial velocity of 52 ft/sec. Its height above the ground after seconds is given by the formula h(t)=-16t^2+52t. What is the maximum height that the ball attains before hitting the ground (feet)?
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Max height is at the vertex.
Value of t at vertex is:
t = -b/(2a)
t = -52/(2(-16))
t = -52/(-32)
t = 1.625
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Height at this time is:
h(t)=-16t^2+52t
h(1.625)=-16(1.625)^2+52(1.625)
h(1.625)= 42.25 feet
