Abcdefg A.
asked • 06/26/16Calculus Question: relationship between height and velocity
Determine the height from which an object must be dropped so that it will have the same speed in ft /?s when it lands as the height in feet from which it was dropped.
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Degonimia H. answered • 07/01/16
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Alan G. answered • 06/26/16
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Alan G.
Dear Abracadabra,
Let s(t) = -16t2 + h be the height of the object t seconds after it is dropped from a height of h feet.
Its velocity at time t will be s′(t) = -32t. Here is a sketch of steps needed to solve this problem.
1) Determine how long it will take the object to reach the ground. This is accomplished by solving s(t) = 0 for t.
2) Set |v(t)| = h and solve for t. I use absolute value because speed is the absolute value of velocity.
3) Equate the times from steps 2) and 3) and solve for h.
Here are the details.
1) -16t2 + h = 0 ⇒ t = (√h)/4
2) 32t = h ⇒ y = h/32
3) (√h)/4 = h/32 ⇒ 8√h = h ⇒ 64h = h2 ⇒ 0 = h2 - 64h = h(h - 64) ⇒ h = 0 or h = 64.
The correct answer is 64 feet.
Report
07/02/16
Alan G.
I resubmitted the solution after realizing it was incorrectly posted yesterday. Sorry for the problem!
Report
07/03/16
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Alan G.
Let s(t) = -16t2 + h be the height of the object t seconds after it is dropped from a height of h feet.
Its velocity at time t will be s′(t) = -32t. Here is a sketch of steps needed to solve this problem.
1) Determine how long it will take the object to reach the ground. This is accomplished by solving s(t) = 0 for t.
2) Set |v(t)| = h and solve for t. I use absolute value because speed is the absolute value of velocity.
3) Equate the times from steps 2) and 3) and solve for h.
Here are the details.
1) -16t2 + h = 0 ⇒ t = (√h)/4
2) 32t = h ⇒ y = h/32
3) (√h)/4 = h/32 ⇒ 8√h = h ⇒ 64h = h2 ⇒ 0 = h2 - 64h = h(h - 64) ⇒ h = 0 or h = 64.
The correct answer is 64 feet.
07/03/16