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If a ball is thrown vertically upward from the roof of 32 foot building with a velocity of 112 ft/sec, its height after t secon...Asked by Jin
If a ball is thrown vertically upward from the roof of 32 foot building with a velocity of 80 ft/sec, its height after t seconds is s(t)=32+80t–16t2. What is the maximum height the ball reaches?
What is the velocity of the ball when it hits the ground (height 0)?
for max height I put 80-32t
and for velocity i put 32...which are wrong..
What is the velocity of the ball when it hits the ground (height 0)?
for max height I put 80-32t
and for velocity i put 32...which are wrong..
Answers
Answered by
drwls
The building height is not the velocity.
For the max height, rewrite
s(t) = -16t^2 +80t +32 as
-16(t^2 - 5t + 25/4) +132
= -16(t -5/2)^2 + 132
The highest possible value of the function is 132 feet, and it occurs when t = 2.5 seconds.
For the velocity when it hits the ground, first solve for t when s = 0.
Then use the equation for velocity vs time:
V = 80 - 32 t ft/s
For the max height, rewrite
s(t) = -16t^2 +80t +32 as
-16(t^2 - 5t + 25/4) +132
= -16(t -5/2)^2 + 132
The highest possible value of the function is 132 feet, and it occurs when t = 2.5 seconds.
For the velocity when it hits the ground, first solve for t when s = 0.
Then use the equation for velocity vs time:
V = 80 - 32 t ft/s
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