Slove the math step by step

Question

A rock is dropped from a 200 m –high cliff. How long does it take to fall (a) the first 100 m and (b) the last 50 m.

Answer 1

Let:

y = some arbitrary distance between height of cliff and the ground

v1 = initial velocity (0m/s)

t = the time it takes for the rock to drop distance "y"

g = gravitational acceleration (9.81m/s^2)

Equation to use:

y2 = y1 + v1*t + (1/2)a*t^2

Define coordinate system:

Define the down direction as positive. Set the top of the cliff as the zero point (y1 = 0)

y1 = 0, y2 = y, v1 = 0, v2 = v2, t = t

Let's take a look of how time "t" varies as we adjust "y"

Plug in these variables into the equation:

y = 0 + 0 + (1/2)(g)(t^2)

Algebraically rearrange the equation to solve for t

t = sqrt( 2*y/g )

For part (a) y = 100m, so directly plug that into the equation for time:

t = sqrt( 2*100m / 9.81m/s^2) = 4.51s

For part (b), let's figure out the difference in the two times between the 50m point and the ground.

50m point: 50m above the ground is 150m displacement from the top of the cliff. So let y = 150m

t = sqrt( 2*150m / 9.81m/s^2) = 5.53s

This is time from the top of the cliff to the 50m point

ground: let y = 200m

t = sqrt( 2*200m / 9.81m/s^2) = 6.39

This is time from the top of the cliff to the ground

difference in times: 6.39s - 5.53s = 0.855s

This is time from the 50m point to the ground. Answer for part B