Question

A stone is thrown straight up from the edge of a roof, 700 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?

Answer 1

Since, acceleration(a) = dv/dt

given, a = -g = -32 ft/s^2

then,

dv = -32*dt

By integrating it,

v(t) - v(0) = -32*(t - 0)

v(t) = v(0) - 32*t

here, v(0) = initial speed = 10 ft/s

So, v(t) = 10 - 32*t eq(1)

also, v(t) = dx/dt

dx = (10 - 32*t)*dt

Again by integrate it,

x(t) - x(0) = (10*t - 32*t^2/2) - (10*0 - 32*0)

x(t) = x(0) + 10*t - 16*t^2

given, x(0) = initial height = 700 ft

So,

x(t) = 700 + 10*t - 16*t^2

A.

at t = 5 seconds,

x(5) = 700 + 10*5 - 16*5^2

x(5) = 350 m

B.

Let, 't' second after stone hit the ground.

at ground, x(t) = 0

So,

0 = 700 + 10*t - 16*t^2

16*t^2 - 10*t - 700 = 0

By solving above quadratic equation,

t = 6.93 s

C.

from eq(1),

v(t) = 10 - 32*(6.93)

v(t) = -211.76 ft/s (here, negative signt means that velocity is in downward direction.)

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