(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A projectile starts from rest and moves 4.7 m down a frictionless ramp inclined at 21 degrees with the horizontal. the acceleration due to gravity is 9.8 m/s^2

what what speed will it leave the ramp ?

what will the range of the projectile if the bottom of the ramp is 2.1 m above the ground

2. Relevant equations

V=Vi + at

delta X = Vi*t+1/2 a*t^2

Vf^2 - Vi^2 = 2*a*deltaX

3. The attempt at a solution

Vertical,

Viy=0

ay=-9.8

Vfy= - Vf sin 21

delta y= -4.7 sin 21

horizontal,

Vix= ( i thought it to be zero at first but it cant be zero because then the delta x comes out to be zero. So im confused at this one )

ax=0

Vfx=Vf cos 21

delta x= 4.7 cos 21

so i used the third law and set it up as follows,

Vfy^2 = 2 (-9.8) (4.7 sin 21)

then the idea is to set the answer = Vf sin 21 and solve for Vf

it comes out to be 16.0329

but this is unfortunately not the right answer. can anyone help ?

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# Projectile and ramp problem

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