Question

A boulder with a mass of 2,500 kg on a ledge 200 m above the ground falls. Estimate the speed of the boulder just before it hits the ground.

Answer 1

using vf^2= vi^2 +2gh, theres no initial velocity so this can be rewritten to
 v^2 = 2gh or v = sqrt(2gh)
v = sqrt(2*200m*9.81m/s^2) 
v = 62.6418 m/s
v = 62.6 m/s

Answer 2

If a boulder with a mass of 2,500 kg on a ledge 200 m above the ground falls, the speed of the boulder just before it hits the ground is 62.60m/s

The formula for calculating the speed of boulder will be expressed as:

$v_f^2=v_i^2+2gh$

$v_f \ and \ v_i$ are the final and initial velocities respectively

g is the acceleration due to gravity

h is the height reached

Given the following parameters:

$m=2500\\h=200m$

Substitute the given parameters into the formula as shown:

$v_f^2=0+2(9.8)(200)\\v_f^2=3920\\v_f=\sqrt{3920}\\ v_f=62.60m/s$

Hence the speed of the boulder just before it hits the ground is 62.60m/s

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