Rachel is at window from11th floor of building, bucket passes by. notes it hits ground 1.5s later. What's the height from bucket was dropped- each floor 5m high, window in middle of the 11th floor.

Answer 1

In some parts of the world the ground floor counts as floor number 0.

There the middle of the 11th floor is at height 11×5 + 2.5 m from the ground.

In other parts of the world the ground floor counts as floor number 1.

There the middle of the 11th floor is at height 10×5 + 2.5 m from the ground.

I do not know how you count it, so I will leave the height of Rachel's

window unspecified, denoted by h.

Let

h be the height of Rachel's window,

t denote time with t=0 being the instant the bucket is dropped,

x(t) be the height of the bucket at time t,

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

t₁ be the time instant the bucket passes Rachel's window,

t₂ = t₁+1.5 be the time instant the bucket hits the ground,

g = -9.81 m/s² be gravitational acceleration.

I am counting upward direction as positive, therefore the heights h and x(t) are positive.

I am counting downward direction as negative, therefore g is negative.

We are given

v(0) = 0, i.e., the bucket is dropped, not thrown down (1)

x(t₁) = h i.e., at time t₁ it passes Rachel's window (2)

x(t₂) = 0 i.e., at time t₂ it hits the ground (3)

We are to calculate x(0).

Since this is a calculus question, not physics question, I will derive the equations of motion.

The bucket is under constant acceleration g.

Its velocity v(t) is the integral of acceleration, so

v(t) = ∫gdt = gt + V, for some constant V

We calculate V from the constraint (1)

0 = g×0 + V

V = 0

So

v(t) = gt

Position is integral of velocity, so

x(t) = ∫v(t)dt = ∫gtdt = gt²/2 + C, for some constant C.

At time t=0

x(0) = C

So we will write

x(t) = gt²/2 + x(0)

Now we can substitute the expression for x(t) into (2) and (3)

getting two equations with two unknowns -- x(0) and t₁.

gt₁²/2 + x(0) = h

gt₂²/2 + x(0) = 0

Subtract the two equations

gt₁²/2 - gt₂²/2 = h

gt₁² - g(t₁ + 1.5)² = 2h

gt₁² - gt₁² - 3gt₁ - 2.25g = 2h

t₁ = -(2h + 2.25g)/3g

Once you substitute the known g and your value of h, you get t₁.

(Remember that g is negative.)

Then you can calculate the desired

x(0) = h - gt₁²/2