the height h (in feet) relative to the point of release of an object t sec after it is thrown straight…

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 32 ft/sec is given by the equation h = 32t - 16t². 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 seconds. (simplify your answer. type an integer or a decimal. round to the nearest tenth.)
Answer
Explanation:
Step1: Set height equal to initial height
When the object returns to the original height, $h = 0$. So we set the equation $32t-16t^{2}=0$.
Step2: Factor out the common factor
Factor out $16t$ from the left - hand side of the equation: $16t(2 - t)=0$.
Step3: Use the zero - product property
If $ab = 0$, then either $a = 0$ or $b = 0$. So $16t=0$ or $2 - t=0$. The solution $t = 0$ corresponds to the time of release. Solving $2 - t=0$ gives $t = 2$.
Answer:
2