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 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

Hi James!

For this problem, since the stone is subject to a constant acceleration due to gravity for part A we can employ the kinematic equation:

Δy = V₀t + (1/2)at²

where Δy is the vertical displacement from the rooftop (we will call the upwards direction positive), V₀ is the initial velocity of the stone (12 feet per second up), a is the constant gravitation acceleration (32 feet per second squared down), and t is the time since release (5 seconds). In this case, we can simply plug the values into the equation, making sure to use the correct sign, and arrive at the answer:

Δy = V₀t + (1/2)at² = (12 ft/s)(5 s) + (1/2)(-32 ft/s²)(5 s)²

Δy = -340 ft

This indicates that the stone is 340 feet below where it was released. Since the stone was released from a roof 650 feet above the ground the stone is 310 feet (650 ft - 340 ft) above the ground after 5 seconds.

For part B, you could solve the problem in two different ways. The first would be to use the fact that Δy = -650 feet when the stone hits the ground and solve for t in the equation used in part A using the quadratic formula. The second method involves first finding the velocity when the stone hits the ground using the kinematic equation:

V_f² = V₀² + 2aΔy

where V_f is the final velocity when the stone hits the ground. Then use the kinematic equation:

V_f = V₀ + at

and solve for t, the time it takes to reach V_f, which is also the time it takes to reach the ground. In this case, because part C asks for the velocity when the stone reaches the ground, we might as well use method 2, so we can kill two birds with one stone (pun intended). For the first part, V₀ = 12 ft/s, a = -32 ft/s², and Δy = -650 ft:

V_f² = V₀² + 2aΔy = (12 ft/s)² + 2(-32 ft/s²)(-650 ft) = 41,744 ft²/s²

V_f = ±204.3 ft/s

The square root gives a plus or minus but using the fact that we know the stone must be moving downwards when it hits the ground, we can say that the final velocity is -204.3 ft/s or 204.3 ft/s downwards. Using this value we can solve for the time it took to reach the ground:

V_f = V₀ + at

t = (V_f - V₀)/a = ((-204.3 ft/s) - (12 ft/s))/(-32 ft/s²)

t = 6.76 seconds

The answer to part B is that the stone takes 6.76 seconds to reach the ground and, conveniently, we also found the answer to part C, which is 204.3 ft/s towards the ground.

I hope this helps!

Seth

Answer 2

Let

g = -32 ft/s² be gravitational acceleration,

v(t) be the velocity of the stone at time t,

h(t) be the stone's height (as measured from ground) at time t,

t = 0 be the instant the stone is thrown.

We are given

h(0) = 650 ft,

v(0) = 12 ft/s

Given that this is a calculus class, not physics class, I assume you are not

expected to know the formulas for v(t) and h(t); instead you are supposed to derive them.

So here is the derivation.

The definition of velocity v is

v = ds/dt

The definition of acceleration g is

g = dv/dt

Therefore we can calculate v(t), taking advantage of being given g as a constant

v(t) = ∫gdt = gt + v(0)

From that we can calculate s

s(t) = ∫vdt = gt²/2 + v(0)t + s(0)

A.

h(5) = -32×5²/2 + 12×5 + 650 = 310 ft

B.

We need to solve for t the equation

h(t) = 0

-32t²/2 + 12t + 650 = 0

This has two solutions:

1) t = 6.76 s

which is the desired answer.

2) t = -6 s

which is not physically feasible.

It is an answer to a different question:

"At what time would somebody have to throw a stone from ground level so that

at time 0 it reaches the edge of the roof with speed 12 ft/s?"

C.

v(6.76) = -32×6.76 + 12 = -204 ft/s

The velocity is negative because its direction is downward.