Question

A rock is thrown off of a 150 foot diff with an upward velocity of 40 fus. As a result its height after t seconds is given by the formulah(t) = 150 + 40t - 5t^2a) What is its height after 4 seconds?b) What is its velocity after 4 seconds?(Positive velocity means it is on the way up, negative velocity means it is on the way down.)

Answer 1

Given the formula:

`h(t)=150+40t-5t^2`

Let's solve for the following:

• (a). What is its height after 4 seconds?

To find the height after 4 seconds, substitute 4 for t and solve for h(4):

`\begin{gathered} h(4)=150+40(4)-5(4)^2 \\ \\ h(4)=150+160-80 \\ \\ h(4)=230 \end{gathered}`

The height after 4 seconds is 230 feet.

• (b). What is its velocity after 4 seconds?

To find the velocity, we have:

`v(t)=(dh)/(dt)=(d)/(dt)(150+40t-5t^2)`

Now find the derivative:

`v(t)=40-10t`

Substitute 4 for t and solve for v(4):

`\begin{gathered} v(4)=40-10(4) \\ \\ v(4)=40-40 \\ \\ v(4)=0\text{ m/s} \end{gathered}`

Therefore, after 4 seconds, the velocity is 0 m/s.

ANSWER:

a) 230 ft

b) 0