Question

If a building is 44 m tall, how long would it take to fall off it?

Answer 1

2.995 seconds

Step-by-step explanation:

We can use this kinematics equation to evaluate time.

`y=v_0t-(1)/(2)gt^2`

Lets solve for
`t.`

Combine
`(1)/(2)` and
`gt^2`.

`y=v_0t-(gt^2)/(2)`

Subtract
`y` from both sides.

`0=v_0t-(gt^2)/(2)-y`

`0=-(gt^2)/(2) -v_0t-y`

Multiply both sides of the equation by -1.

`0=(gt^2)/(2) -v_0t+y`

Multiply both sides of the equation by 2.

`0=gt^2 -2v_0t+2y`

Use the quadratic formula to solve for t.

`(-b+√(b^2-4ac) )/(2a)`

`(-b-√(b^2-4ac) )/(2a)`

`a=-g\\b=2v_0\\c=-2y`

Solution 1 Steps

`t=\frac{-2v_0+\sqrt{(2v_0)^(2) -4*-g*-2y} }{2*-g}`

`t=(-2v_0+√(4v_0^2 -4*-g*-2y) )/(2*-g)`

`t=(-2v_0+√(4v_0^2 -8gy) )/(-2g)`

Solution 2 Steps

`t=\frac{-2v_0-\sqrt{(2v_0)^(2) -4*-g*-2y} }{2*-g}`

`t=(-2v_0-√(4v_0^2 -4*-g*-2y) )/(2*-g)`

`t=(-2v_0-√(4v_0^2 -8gy) )/(-2g)`

One of these solution will most likely lead to a negative answer. The solution that gives a positive answer is correct. Lets enter our values into both equations to see which one is correct.

We can assume the initial velocity is 0.

We are given

`v_0=0\\g=-9.81\\y=44`

Lets try the first solution.

`t=(-2*0+√((4*0^2) -(8*-9.81*44)) )/(-2*-9.81)`

`t=(0+√(0 -(8*-9.81*44)) )/(-2*-9.81)`

`t=(0+√(0 --3453.12) )/(-2*-9.81)`

`t=(√(3453.12) )/(19.62)`

`t=2.995`

Our first solution was positive so there is no need to check the second solution.