Calculus Integration Problem 2

Question

A stone is dropped from the edge of a roof, and hits the ground with a velocity of −180 feet per second. Assume the acceleration due to gravity is -32 feet per second squared. How high (in feet) is the roof?

 

Could someone explain how to go about this?

Answer 1

Hi Anna,

 

We can use some of what I talked about in your previous question for this problem (so read my answer to that one first!!).

 

We know

a(t)=-32, so

v(t)=int -32 dt, which  means

v(t)=-32t+C.

 

To find C in this problem, we need to know the velocity of the stone when t=0. Well, in this problem, it says the stone is dropped, which means the velocity when t=0 is actually 0. Plugging that in, we get

0=v(0)=-32(0)+C, so C=0. 

So v(t)=-32t.

 

Again, from the last problem I helped you with,

s(t)=int v(t)dt, so

s(t)=int -32t dt, which means

s(t)=-16t²+K.

 

When t=0, the stone should be the same height as the building, so the position function should be equal to the building height. That tells us that

s(0)=-16(0)+K is the height of the roof. SO, the height of the roof is equal to K, and we need to figure out how to find K!

 

We haven't used all of the information from the problem yet. It says the stone hits the ground at a velocity of -180 ft/sec. But when does the stone hit the ground? Well, just like in the last problem, to find when the stone hits the ground, we need to set the position function equal to 0:

0=s(t)=-16t²+K.

Solving this for t, we get

16t²=K

t²=K/16, so

t=sqrt K/4 (we don't need to use plus or minus because time has to be positive).

 

Now, we can plug that in to our velocity function. When it hits the ground, we know the velocity is -180, so

 

-180=v(sqrtK/4)=-32(sqrtK/4)

 

This simplifies to

-180=-8 sqrtK, so

sqrtK=45/2. Square both sides to find K, which is the height of the roof!

 

Let me know if you need any more help or if anything wasn't clear!

Erika