A particular object is projected vertically upwards from the ground.After t- seconds it’s height is h, m, where h = 50t-5t^2 Find the greatest height reached and the times when the particle is 120m highland it’s speed st these times?

1 Answer
Noah G
Mar 20, 2018

The greatest height will be the absolute max of the given function. Taking the derivative we get

#h'(t) = 50 - 25t#

We will have a critical point when #h'(t) = 0#.

#0 = 50 - 25t#

#25t = 50#

#t = 2#

Since this is a parabola that opens downwards, our singular critical point is certainly a maximum. At time #t = 2# seconds, the object will have height given by #h(2) = 50(2) - 5(2)^2 = 100 - 20 = 80 m#

I'm not sure about your second question--please clarify.

Hopefully this helps!

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