Question

an object is dropped from a tower, 163 ft above the ground. the objects height above ground t sec into the fall is s = 163 - 16t². a. what is the objects velocity, speed, and acceleration at time t? b. about how long does it take the object to hit the ground? c. what is the objects velocity at the moment of impact? the objects velocity at time t is - 32t. the objects speed at time t is 32t ft/sec. the objects acceleration at time t is - 32 ft/sec². (simplify your answer.) it takes 3.2 sec for the object to hit the ground. (round to the nearest tenth.) the objects velocity at the moment of impact is ft/sec. (round to the nearest tenth.)

Explanation:

Step1: Recall velocity - derivative of position

Velocity $v(t)$ is the derivative of the position function $s(t)$. Given $s(t)=163 - 16t^{2}$, using the power - rule $\frac{d}{dt}(x^{n})=nx^{n - 1}$, we have $v(t)=s^\prime(t)=\frac{d}{dt}(163-16t^{2})=-32t$.

Step2: Recall speed - absolute value of velocity

Speed is the absolute value of velocity. So, speed $=|v(t)| = |-32t|=32t$ (since $t\geq0$ in this context of time of fall).

Step3: Recall acceleration - derivative of velocity

Acceleration $a(t)$ is the derivative of the velocity function. Since $v(t)=-32t$, then $a(t)=v^\prime(t)=\frac{d}{dt}(-32t)=-32$.

Step4: Find time to hit the ground

When the object hits the ground, $s(t) = 0$. So, we set $163-16t^{2}=0$. Then $16t^{2}=163$, and $t^{2}=\frac{163}{16}$. Solving for $t$, we get $t=\sqrt{\frac{163}{16}}\approx3.2$ s.

Step5: Find velocity at impact

We know the time of impact $t\approx3.2$ s. Substitute $t = 3.2$ into the velocity function $v(t)=-32t$. So, $v(3.2)=-32\times3.2=-102.4$ ft/sec.

Answer:

a. Velocity: $v(t)=-32t$ ft/sec, Speed: $32t$ ft/sec, Acceleration: $a(t)=-32$ ft/sec²
b. $t\approx3.2$ s
c. $v\approx - 102.4$ ft/sec