At time t = 0, a projectile was fired upward from an initial height of

Question

At time t = 0, a projectile was fired upward from an initial height of 10 feet. Its height after t seconds is given by the function \(h(t) = p −10 (q − t)^2\), where p and q are positive constants. If the projectile reached a maximum height of 100 feet when t = 3, then what was the height, in feet, of the projectile when t = 4 ?

(A) 62
(B) 70
(C) 85
(D) 89
(E) 90

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Official Answer

E

(A) 62
(B) 70
(C) 85
(D) 89
(E) 90

Are You Up For the Challenge: 700 Level Questions

PyjamaScientist · Answer

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jnmudd15 · Answer

At max height, why did you cross out the -10(q-t)^2 term? I understand that at the max height, velocity = 0. Did you assume that that was the velocity term? I don't understand how that conclusion was determined. Thanks in advance.

stne · Answer

Hi ,

This expression  p - 10(q-t)^2  will be maximum at max height. 
or for the above  expression to be maximum   -10(q-t)^2  will have to be zero.Because if  -10(q-t)^2  has any other value other than 0 then this expression will always reduce the value of  p.

Hence to get max value of  p  at max height, we need to take  - 10(q-t)^2 = 0

Hope it helps.

At time t = 0, a projectile was fired upward from an initial height of

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At time t = 0, a projectile was fired upward from an initial height of

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