Asked by michael
If a ball is thrown vertically upward from the roof of 64 foot building with a velocity of 64 ft/sec, its height after t seconds is s(t)=64+64t–16t2 . What is the maximum height the ball reaches? What is the velocity of the ball when it hits the ground (height 0 )?
Answers
Answered by
jim
Differentiate to find the max height.
Velocity is 64 - 32t.
Max is at s'(t) = 0, when velocity is zero
s'(t)= 64 - 32t = 0
so it peaks at t=2. From that you can get the height.
You now need to find s(t) = 0, so solve 64+64t–16t2 = 0
That will give you the time, and from that the velocity function 64 - 32t will give you the answer to the second part
Velocity is 64 - 32t.
Max is at s'(t) = 0, when velocity is zero
s'(t)= 64 - 32t = 0
so it peaks at t=2. From that you can get the height.
You now need to find s(t) = 0, so solve 64+64t–16t2 = 0
That will give you the time, and from that the velocity function 64 - 32t will give you the answer to the second part
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