the height of an object at time t ≥ 1 is given by h(t)=t² - 16/t + 15. what is the velocity of the object at…

Explanation:

Step1: Recall velocity - height relationship

Velocity $v(t)$ is the derivative of height $h(t)$. Given $h(t)=t^{2}-\frac{16}{t}+15=t^{2}-16t^{- 1}+15$.

Step2: Differentiate $h(t)$

Using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=h^\prime(t)=2t+16t^{-2}=2t+\frac{16}{t^{2}}$.

Step3: Evaluate $v(t)$ at $t = 3$

Substitute $t = 3$ into $v(t)$: $v(3)=2\times3+\frac{16}{3^{2}}=6+\frac{16}{9}=\frac{54 + 16}{9}=\frac{70}{9}\approx7.778$.

Answer:

B. 7.778