Asked by Sam

A child's toy consists of a block that attaches to a table with a suction cup, a spring connected to that block, a ball, and a launching ramp. (Intro 1 figure) The spring has a spring constant k, the ball has a mass m, and the ramp rises a height y above the table, the surface of which is a height H above the floor.

Initially, the spring rests at its equilibrium length. The spring then is compressed a distance s, where the ball is held at rest. The ball is then released, launching it up the ramp. When the ball leaves the launching ramp its velocity vector makes an angle theta with respect to the horizontal.

Throughout this problem, ignore friction and air resistance.

Calculate v_r, the speed of the ball when it leaves the launching ramp.
Express the speed of the ball in terms of k, s, m, g, y, and/or H.

I thought it's supposed to be
sq(2mgyH)/m

wouldn't it just have potential and kinetic energy? apparently that's incorrect.
Answered by drwls
You seem to be neglecting the stored energy in the compressed spring, which is very important
Answered by Sam
Would it be sq(2ks^2/m)?

Answered by Ann
sq((ks^2-mgy)/m)
Answered by ???
sq((ks^2/m)-2gy)
Answered by yo
answer does not include h or y.
Answered by Paul
sqrt(ks^2-2mgy/m)
Answered by Munir
sqrt(ks^2-2mgy)/m
Answered by CHEater
ITS DEFF
sqrt(ks^2-2mgy/m)
Paul ALL DAY
Answered by SANDY DAMICO
/sqrt((ks^2 -mgy)/m)
Answered by Bruce Wayne
E1=E2
E1=(ks^2)/2
E2=mgh+(mv^2)/2
(ks^2)/2=mgh+(mv^2)/2
2*[(ks^2)/2-mgh]/m=v^2
sq[(ks^2-2mgh)/m]=v
Answered by Naj
wq[(ks^2-2mgy)/m]=v
its y not h
Answered by KAT
REAL FINAL ANSWER: SQUARE ROOT OF (ks^2/m+2gH)
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