Calculus Integration Problem

Question

A stone is thrown straight down from the edge of a roof, 900 ft above the ground, at a speed of 7 ft/sec.

 

A. Given that the acceleration due to gravity is -32 ft/sec2, how high is the stone 2 seconds later?

B. At what time does the stone hit the ground?

C. What is the velocity of the stone when it hits the ground?

 

Would someone be able to explain this to me?

Answer 1

Hi Anna!

 

This sounds like a problem where we start with acceleration, and then integrate to get the velocity and position functions so answer the questions. So, let's do that!

 

The question tells you that the acceleration function is a constant -32, so

a(t)=-32

 

You (hopefully) learned in class that the antiderivative of acceleration is the velocity function, so (if int stands for integral)

v(t)=int (-32)dt, which gives us

v(t)=-32t+C.

 

To find C, we need to know one value of v(t), or the velocity. Good news is that we do! The problem tells us the stone is thrown at a speed of 7 ft/sec, so that means that when t=0, the velocity is -7 (since it is thrown down, the velocity will be negative, just like acceleration).

 

If we plug t=0 into the velocity function, we get

v(0)=-7=-32(0)+C, so C=-7.

Now you know your velocity function:

v(t)=-32t-7

 

To get your position function, you (again, hopefully) learned in class that the antiderivative of velocity is the position function. So, the position function (which is usually called s(t)) is:

s(t)=int v(t)dt

s(t)=int (-32t-7)dt

s(t)=-16t²-7t+K.

 

To find the constant K, we need one value of the position function, which we have! The stone started at 900 ft above ground, so when t=0, s=900. Plug that into your position function to get

900=s(0)=-16(0)-7(0)+K, so K=900.

 

So, your position function is 

s(t)=-16t²-7t+900

 

We can use these to solve your questions.

 

A) To find how high the stone is 2 seconds later, just plug t=2 into your position function s(t)

B) To find when the stone hits the ground, you want to know when the position is 0, so set 

s(t)=0, or -16t²-7t+900=0 and solve for t using the quadratic formula

C) When you find the value of t corresponding to when the stone hits the ground in part B, you can plug that in to your velocity function v(t) to find the velocity of the stone when it hits the ground.

 

Let me know if that helps! If you'd like further help, we can always set up a private tutoring session!

Erika