Question

90 points whats the equation for how fast something will be traveling before it hits the ground. How to do the equation PLEASE EXPLAIN here is an example problem
A 0.05 kg-coin is dropped from the top of a skyscraper that is 100 meters high. How fast is the coin traveling just before it hits the ground?
use GPE=M*g*h
KE= 0.5*Mass *velocity squared

Answer 1

given GPE=M*g*h

KE= 0.5*Mass *velocity squared

something falls n before it hits the ground,

by conservation of energy

GPE loss = KE gain

e.g. A 0.05 kg-coin is dropped from the top of a skyscraper that is 100 meters high.

so M=0.05, g=9.8m/s^2, h=100m

GPE loss=M*g*h=0.05*9.8*100

=49J

KE gain=0.5*Mass *velocity squared=GPE loss

0.5*0.05*velocity squared=49

velocity squared=49/0.5/0.05=1960

velocity=sqrt(1960)=44.27m/s

Answer 2

I will solve it using a different set of equations from kinematics. The same principles are the same but the equations look different. Specifically

final velocity^2 = initial velocity^2 + 2(acceleration)(displacement)

In this example, the coin has no initial velocity, acceleration is by gravity and the displacement is 100m

final velocity^2 = 0 + 2(9.8)(100)=1960

final velocity = (1960)^(1/2) = 44.3m/s