A stone is thrown straight up from the edge of a roof, 650 feet above the ground, at a speed of 12 feet per second.

Question

A stone is thrown straight up from the edge of a roof, 650 feet above the ground, at a speed of 12 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?   

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

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

I am not sure where to start...idk if 12t-32t^2 is my height equation and so 12-32t=0 give mes .375 for t and I plug that value into 12t-3t^2???

Answer 1

Bri

acceleration = dv/dt

Lets assume positive values are upwards, negative downwards.

So a(0) = acceleration at time zero = - 32

v(0) = speed at time zero = + 12

s(0) = distance above ground at time zero = + 650

dv/dt = -32 as dv/dt = acceleration

dv = -32 dt

Integrate both sides

v = -32 t + C

v(0) = 12, so that means C = 12

So v = -32t + 12

v = ds/dt

ds/dt = -32t + 12

ds = (-32t + 12) dt

Integrate both sides

s = -16t² + 12t + C

s(0) = 650, so C = 650

s = -16t² + 12t + 650

So now we have expressions for a(t) = -32, v(t) and s(t)

We can use these

to answer the questions now.

For (a), you need s(2). Exercise for you.

For (b) you need to find t when s(t) = 0.

For (c), you need to find v(t) for the value of t you found in (b).

Will leave these for you as exercises - ADD A COMMENT IF YOU NEED FURTHER HELP

Mike