Question

A rock is thrown off of a 120 foot cliff with an upward velocity of 20 ft/s. As a result its height after t seconds is given by the formula:h(t) = 120 + 20t - 5t^2What is its height after 2 seconds?___What is its velocity after 2 seconds?____(Positive velocity means it is on the way up, negative velocity means it is on the way down.)

Answer 1

We are given that the height of a rock in terms of the time is given by the following equation:

`h\mleft(t\mright)=120+20t-5t^2`

We are asked to determine the height after two seconds. To do that we will substitute in the equation the value of "t = 2s", like this:

`h(2)=120+20(2)-5(2)^2`

Solving the operations we get:

`h(2)=140`

Therefore, the height after 2 seconds is 140 ft.

Now, to determine an equation for the velocity we will determine the derivative with respect to the time of the equation for the height.

`(dh)/(dt)=(d)/(dt)(120+20t-5t^2)`

Now, we distribute the derivative:

`(dh)/(dt)=(d)/(dt)(120)+(d)/(dt)(20t)-(d)/(dt)(5t^2)`

For the first derivative we will use the following rule:

`(d)/(dt)(a)=0`

Where "a" is a constant. Applying the rule we get:

`(dh)/(dt)=(d)/(dt)(20t)-(d)/(dt)(5t^2)`

For the second derivative we will use the following rule:

`(d)/(dt)(at)=a`

Where "a" is a constant. Applying the rule we get:

`(dh)/(dt)=20-(d)/(dt)(5t^2)`

For the last derivative we will use the following rule:

`(d)/(dt)(at^n)=\text{nat}^(n-1)`

Applying the rule we get:

`(dh)/(dt)=20-10t`

Since the derivative of the position with respect to time is the velocity we have:

`(dh)/(dt)=v=20-10t`

Now, we substitute the value of "t = 2s":

`v=20-10(2)`

Now, we solve the operations:

`\begin{gathered} v=20-20 \\ v=0 \end{gathered}`

Therefore, the velocity after 2 seconds is 0.