Question 825435: Michael throws a baseball into the air with an initial vertical velocity of 180 feet per second from a 100 foot cliff.
A: Write the height of the baseball as a function of time.
B: When will the baseball be 200 feet high?
C: When will the baseball reach its maximum height?
D: How high is the baseball after 3 seconds?
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! ---
equation of ballistic motion:
y(t) = (1/2)gt^2 + v0t + y0
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where y(t) is height at time t
g = acceleration due to gravity
v0 = initial velocity
y0 = initial height
t = time
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y(t) = (1/2)-32t^2 + 180t + 100
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answer A:
y(t) = -16t^2 + 180t + 100
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setup for B:
200 = -16t^2 + 180t + 100
-16t^2 + 180t - 100 = 0
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the above quadratic equation is in standard form, with a=-16, b=180, and c=-100
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
-16 180 -100
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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the quadratic has two real roots at:
t = 0.586088907
t = 10.6639111
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the quadratic vertex is a maximum at: ( t= 5.625, y(t)= 406.25 )
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answer B:
the ball is at 200 feet twice:
t = 0.586088907 seconds
t = 10.6639111 seconds
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answer C:
at t= 5.625 seconds the ball reaches its maximum height of 406.25 feet
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answer D:
y(3) = -16*3^2 + 180*3 + 100
y(3) = 496 feet
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