Question

an object is dropped from a tower, 165 ft above the ground. the objects height above ground t sec into the fall is s = 165 - 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 - 102.4 ft/sec (round to the nearest tenth.)

Explanation:

Step1: Find velocity

Velocity $v$ is the derivative of position $s$. Given $s = 165-16t^{2}$, using the power - rule $\frac{d}{dt}(x^{n})=nx^{n - 1}$, we have $v=\frac{ds}{dt}=\frac{d}{dt}(165-16t^{2})=-32t$.

Step2: Find speed

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

Step3: Find acceleration

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

Step4: Find time to hit the ground

When the object hits the ground, $s = 0$. So, we set $165-16t^{2}=0$. Then $16t^{2}=165$, and $t^{2}=\frac{165}{16}$. Solving for $t$, we get $t=\sqrt{\frac{165}{16}}\approx3.2$ s.

Step5: Find velocity at impact

We substitute the time $t\approx3.2$ s into the velocity formula $v=-32t$. So, $v=-32\times3.2=-102.4$ ft/s.

Answer:

The object's velocity at time $t$ is $-32t$ ft/s.
The object's speed at time $t$ is $32t$ ft/s.
The object's acceleration at time $t$ is $-32$ ft/s².
It takes approximately $3.2$ s for the object to hit the ground.
The object's velocity at the moment of impact is approximately $-102.4$ ft/s.