SOLUTION: A building is 48 ft. tall. From the top of this building a ball is thrown straight up with an initial velocity of 32 ft. per second, and falls to the ground below. The equation t
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Question 84366: A building is 48 ft. tall. From the top of this building a ball is thrown straight up with an initial velocity of 32 ft. per second, and falls to the ground below. The equation that gives the height "s" of the ball "t" seconds after it is thrown is s = -16t^2 + 32t + 48. Find the maximum height reached by the ball and the time it takes for the ball to hit the ground. Answer by funmath(2933) (Show Source):
You can put this solution on YOUR website! A building is 48 ft. tall. From the top of this building a ball is thrown straight up with an initial velocity of 32 ft. per second, and falls to the ground below. The equation that gives the height "s" of the ball "t" seconds after it is thrown is s = -16t^2 + 32t + 48. Find the maximum height reached by the ball and the time it takes for the ball to hit the ground.
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This is a quadratic equation. If you graphed it, you would have a parabola that opened downward. The s value of the vertex of that parabola would be your maximum height.
This quadratic equation is in the form s=at^2+bt+c
a=-16, b=32, and c=48
We can find the t value of the vertex with the formula:
It reaches it's maximum height in 1 s.
Substitute t=1 into the original equation to find the maximum height. The Max height is 64 ft.
The ball is at ground level at the t intercepts, or when s=0:
t+1=0 or t-3=0
t+1-1=0-1 or t-3+3=0+3
t=-1 or t=3
Negative time is meaningless, so the ball hits the ground in t=3 s
Happy Calculating!!!
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