SOLUTION: Alex throws a ball straight up into the air. The equation s = s(t) = –16t2 + 80t + 44 gives the distance s (in feet) the ball is above the ground t seconds after he throws it.
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-> SOLUTION: Alex throws a ball straight up into the air. The equation s = s(t) = –16t2 + 80t + 44 gives the distance s (in feet) the ball is above the ground t seconds after he throws it.
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Question 825404: Alex throws a ball straight up into the air. The equation s = s(t) = –16t2 + 80t + 44 gives the distance s (in feet) the ball is above the ground t seconds after he throws it.
a) How high is the ball two seconds after he throws it?
b) What is the maximum height of the ball from the ground?
c) At what time does the ball hit the ground? Found 2 solutions by TimothyLamb, josmiceli:Answer by TimothyLamb(4379) (Show Source):
You can put this solution on YOUR website! ---
given:
s(t) = -16t^2 + 80t + 44
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answer A:
s(2) = -16*2*2 + 80*2 + 44
s(2) = 140 ft
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s(t) = -16t^2 + 80t + 44
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the above quadratic equation is in standard form, with a=-16, b=80, and c=44
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
-16 80 44
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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the quadratic vertex is a maximum at: ( t= 2.5, s= 144 )
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answer B:
the ball reaches a maximum height of 144 ft
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the quadratic has two real roots at:
t = -0.5
t = 5.5
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the negative root doesn't fit the problem statement, so use the positive root:
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answer C:
the ball hits the ground 5.5 seconds after being thrown
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You can put this solution on YOUR website!
(a)
In 2 sec, the ball is 140 ft high
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(b)
The time , when the
height is maximum is where
Plug this value back into equation to find
The maximum height is 144 ft
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(c)
The ball hits the ground when is zero for
the 2nd time. Find the roots of the equation ( I can't use the positive square root )
The ball hits the ground in 5.5 sec
Here's the plot:
