SOLUTION: A person standing close to the edge on top of a 112-foot building throws a ball vertically upward. The quadratic function h(t) = −16t^2 + 96t + 112 models the ball's height about
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-> SOLUTION: A person standing close to the edge on top of a 112-foot building throws a ball vertically upward. The quadratic function h(t) = −16t^2 + 96t + 112 models the ball's height about
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Question 1175897: A person standing close to the edge on top of a 112-foot building throws a ball vertically upward. The quadratic function h(t) = −16t^2 + 96t + 112 models the ball's height about the ground, h(t), in feet, t seconds after it was thrown.
a) What is the maximum height of the ball?
(Round your answer to the nearest hundredths.)
b) How many seconds does it take until the ball hits the ground? Answer by ikleyn(52772) (Show Source):
You can put this solution on YOUR website! .
A person standing close to the edge on top of a 112-foot building throws a ball vertically upward.
The quadratic function h(t) = −16t^2 + 96t + 112 models the ball's height about the ground, h(t), in feet,
t seconds after it was thrown.
a) What is the maximum height of the ball?
(Round your answer to the nearest hundredths.)
b) How many seconds does it take until the ball hits the ground?
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Part (a)
To find the maximum height, you first find the time to get maximum height
t = - = - = = 3 second.
Then the maximum height is the given function t(t) at t= 3 seconds
h(3) = -16*3^2 + 96*3 + 112 = 256 feet. ANSWER
Part (b)
To find the time when the ball hits the ground, you solve this equation
h(t) = 0, which is -16t^2 + 96t + 112 = 0.
To simplify, divide both sides bu -16
t^2 - 6t - 7 = 0
Factor left side
(t-7)*(t+1) = 0.
Two roots of the quadratic equation are t= 7 and t= -1;
Only positive value t= 7 is the solution.
The ball hits the ground in 7 seconds. ANSWER
