Question

a model rocket is launched vertically upward from the ground with an initial velocity of 32 ft/s. the height of the rocket in feet after t seconds is given by s(t)=-16t² + 32t. when is the rocket at a height of 16 ft? enter an exact answer. provide your answer below: seconds

Explanation:

Step1: Set up the equation

Set $s(t)=16$, so $- 16t^{2}+32t = 16$.

Step2: Rearrange to standard quadratic form

Divide the entire equation by -16 to get $t^{2}-2t=-1$. Then rewrite it as $t^{2}-2t + 1=0$.

Step3: Factor the quadratic equation

The left - hand side factors to $(t - 1)^{2}=0$.

Step4: Solve for t

Take the square root of both sides: $t - 1=0$, so $t = 1$.

Answer:

$1$