Question

A stone is thrown straight up from the edge of a roof, 875 feet above the ground, at a speed of 14 feet per second. A. Remembering that the acceleration due to gravitv is - 32 feet per second squared, how high is the stone 4 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?

Answer 1

we know that ,

acceleration = dv/dt

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

v(0) = speed at time zero = + 14

s(0) = distance above ground at time zero = + 875

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

dv = -32 dt

Integrating both sides:

v = -32 t + C

v(0) = 14, so that means C = 14

So v = -32t + 14

v = ds/dt

ds/dt = -32t + 14

ds = (-32t + 14) dt

Integrating both sides:

s = -16t2 + 14t + C

s(0) = 875, so C = 875

`s = -16t^(2) + 14t + 875\\`

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

for A) s(4)= -16(16) + 14(4)+ 875

s=675

B) find t when s(t)= 0

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