# Math Help - Applications of Derivatives

1. ## Applications of Derivatives

"A stone is thrown straight up from the edge of a roof, 725 feet above the ground, at a speed of 10 feet per second."

"A. Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 2 seconds later?"

I don't fully understand how to do these problems, the basic idea and steps. I have:

dv/dt = -32 I think I have to take the antiderivative of this which would be:

v = -32t + C

I'm not sure what to do afterwards because I don't understand the basic steps behind solving these types of problems. I'm trying to follow the textbook, but it's confusing. There's more parts to the question after this, but I can't study because I don't understand.

2. v = -32t + C.

When t = 0, v = 10. So

v = -32t + 10.

Now v = dx/dt

dx = -32t*dt + 10*dt

Integrate it. When t = 0, x = 725 feet. Then solve for x at t = 2 s.

3. Originally Posted by ImaCowOK
"A stone is thrown straight up from the edge of a roof, 725 feet above the ground, at a speed of 10 feet per second."

"A. Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 2 seconds later?"

I don't fully understand how to do these problems, the basic idea and steps. I have:

dv/dt = -32 I think I have to take the antiderivative of this which would be:

v = -32t + C

I'm not sure what to do afterwards because I don't understand the basic steps behind solving these types of problems. I'm trying to follow the textbook, but it's confusing. There's more parts to the question after this, but I can't study because I don't understand.
Dear ImaCowOK,

Apply the boundry conditions,(when t=0, v=10) then you will be able to find C. Since we want to find ditance the stone has travelled 2 seconds later, we have to get an equation which gives the distance in terms of the time. For that integrate the equation with respect to t again after substituting C. Keep in mind that $v=\frac{ds}{dt}$ where v is the velocity and s is the displacement.

Hope you will be able to solve this problem.