Question

an object is dropped from the top of a cliff 610 meters high. its height above the ground t seconds after it is dropped is 610 - 4.9t². determine its speed 5 seconds after it is dropped. the speed of the object 5 seconds after it is dropped is m/sec (type an integer or a decimal. do not round )

Explanation:

Step1: Recall the relationship between position and velocity

The velocity function $v(t)$ is the derivative of the position - function $s(t)$. Given $s(t)=610 - 4.9t^{2}$, we use the power rule for differentiation. The power rule states that if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$.

Step2: Differentiate the position function

For $s(t)=610 - 4.9t^{2}$, the derivative of the constant 610 is 0 (since the derivative of a constant $C$ is 0, i.e., $\frac{d}{dt}(C)=0$), and the derivative of $-4.9t^{2}$ using the power rule: if $a=-4.9$ and $n = 2$, then $\frac{d}{dt}(-4.9t^{2})=-4.9\times2t=-9.8t$. So, $v(t)=s^\prime(t)=-9.8t$.

Step3: Evaluate the velocity function at $t = 5$

Substitute $t = 5$ into the velocity function $v(t)$. We get $v(5)=-9.8\times5=-49$. The speed is the absolute - value of the velocity. So, the speed $|v(5)| = 49$.

Answer:

49