Question

the height h (in feet) relative to the point of release of an object t sec after it is thrown straight upward with an initial velocity of 64 ft/sec is given by ( h = 64t - 16t^{2} ).

a) how long after it is thrown upwards will it take the object to return to the original height at which is was released?
the object will return in ( square ) seconds.
(simplify your answer. type an integer or a decimal. round to the nearest tenth.)

Explanation:

Step1: Set \( h = 0 \)

When the object returns to the original height, \( h = 0 \). So we have the equation \( 0=64t - 16t^{2} \).

Step2: Factor the equation

Factor out \( 16t \): \( 0 = 16t(4 - t) \).

Step3: Solve for \( t \)

Using the zero - product property \( ab = 0\) implies \(a = 0\) or \(b = 0\).
If \(16t=0\), then \(t = 0\) (this is the time of release).
If \(4 - t=0\), then \(t = 4\).

Answer:

\(4\)