Question

(how to solve?)
A projectile is thrown upward so that its distance above the ground after t seconds is h = -11t2 + 352t. After how
many seconds does it reach its maximum height?

Answer 1

try to find dh/dt

Answer 2

then set dh/dt=0 and solve for t

Answer 3

because at maximum height, the velocity is zero

Answer 4

notice that velocity is dh/dt

Answer 5

differentiate the function. dh/dt gives you the velocity. now when the velocity becomes zero, that means the object has already reached whatever maximum height it could, and will now change its direction to head downwards, so, when v=0, height is maximum.

so, you have to find 't' for which dh/dt =0

\[h= -11t^2 +352t\]
\[\frac{dh}{dt} =-22t + 352\]

-22t+352 = 0
or, t = 352/22 = 16 seconds.


so, at time 't' = 16 seconds, the object reaches its maximum height, and its speed becomes '0'.


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