A stone is thrown straight up from the edge of a roof, 850 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 5 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?

Question

A stone is thrown straight up from the edge of a roof, 850 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 5 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?

anonymous · Answer

There are many different ways to do this, depending on what math or physics class you are taking. Are you in calculus or algebra? If you're in physics, what are the pre-requisites?

anonymous · Answer

It is for a calculus class. I have not yet taken physics

anonymous · Answer

Ok, so we want to first find a formula for the position of the rock over time. Have you done anti derivatives yet? 

Anyway, let's start with -32 as our acceleration. 

d^2y/dt^2=-32
dy/dt=-32t+C

Go back to the original problem. 

10=-32(0)+C=>c=10

integrate again

y=-16t^2+10t+K

850=-16(0)+10(0)+k=>k=850

So our equation is -16t^2+10t+850

If you haven't done anti-derivatives, this should look confusing. Have you?

anonymous · Answer

Well I have gotten that far in the problem I even have the correct answer for A now which is 500ft in the air. I am working on part B and I have tried using the quadratic equation to solve for t but the answer I keep getting isn't working out.