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A roller coaster glides from rest from the top of an 80.0 meter hill. What is the speed of the roller coaster at the bottom of the hill?

Respuesta :

laplacianfuncti
So this is the case of energy conversion. From potential to kinetic.
As energy is conserve:
Epi+Eki=Epf+Ekf ; i for initial and f for final. At begining object at rest hence Eki=0 while at final, object on the lowest position, hence epf=0 therefore:
Epi=Ekf
M.g.h=1/2.m.v^2
Assume no mass change during the process, then
G.h=1/2.v^2
Hence v at bottom:
v=sqrt(2×g×h) assume g=10m/s^2
=sqrt(2×10×80)=sqrt(1600)=40m/s
Hence velocity in the bottom is 40 m/s

At the bottom velocity will be v= 39.60 m/s^2

What is kinematics equation of motion?

Equation of motion with respect to space and time , ignoring the cause of that motion is called  kinematics equation of motion

Given :

u=0

s=80m

g=9.8 m/s^2

using equation of motion

v^2-u^2 = 2as      

where

v = final velocity of the object

u = initial velocity of the object

a = acceleration , in this case a= g as body is falling under the action of gravity pull ( g : acceleration  due to gravity)

s= displacement of the body

v^2-u^2 = 2as      

v^2 -0= 2(9.8)(80)

v^2 = 1568

v= 39.60 m/s^2

At the bottom velocity will be v= 39.60 m/s^2

Learn more about kinematics equation of motion:

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