Question

an object is thrown from the top of an 80 - foot building with an initial velocity of 64 foot per second. the height h of the object after t seconds is given by the quadratic h = - 16t²+64t + 80. when will the object hit the ground? the object will hit the ground when the time is seconds.

Explanation:

Step1: Set height equal to 0

When the object hits the ground, $h = 0$. So we set up the equation $-16t^{2}+64t + 80=0$.

Step2: Divide by -16

Divide the entire equation by - 16 to simplify: $t^{2}-4t - 5=0$.

Step3: Factor the quadratic

Factor the quadratic equation: $(t - 5)(t+1)=0$.

Step4: Solve for t

Set each factor equal to zero:
If $t - 5=0$, then $t = 5$.
If $t + 1=0$, then $t=-1$. Since time cannot be negative in this context, we discard $t=-1$.

Answer:

5