An object is launched upward at 45 ft/sec from a platform that is 40 feet high. what is the objects maximum height if the equation of height (h) in terms of time (t) of the object is given by h(t) = -16t2 + 45t + 40?

The maximum height is determined by using the given equation as written below.
     h(t) = -16t² + 45t + 40

We derive the equation in terms of t and equate the differential to zero.
   dh(t) = -32t + 45 
    0 = -35t + 45

Transpose the constant to the other side of the equation.
    35t = 45
Divide the equation by 35, giving us the value of t which is equal to 9/7. 

This means that the maximum height is attained after 9/7 seconds.

Substitute the computed time to the equation for height.

    h(9/7) = -16(9/7)² + 45(97) + 40

The value of h is equal to 71.41 ft. 

ANSWER: 71.41 ft

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