Asked by Anonymous

A particle, starting from point A in the drawing at a height h0 = 3.2 m, is projected down the curved runway. Upon leaving the runway at point B, the particle is traveling straight upward and reaches a height hf = 4.1 m above the floor before falling back down. Ignoring friction and air resistance, find the speed of the particle at point A.
Answered by bobpursley
Total energy at A= total energy at B
mg*3.2+ KE1=mg*4.1 + 0
solve for KE1, then solve for v1 from

v1= sqrt (KE1/mg)
Answered by notbobpursley
Total Energy at A = Total energy at B

1/2mv^2(A) + mg3.2 = 1/2mv^2(B) + mg4.1
Because at the peak of point B all energy is PE, we can remove 1/2mv^2 (B) from the equation.

1/2mv^2(A) + mg3.2 = mg4.1
because m appears in every term it can be removed.

1/2 v^2 (A) + g3.2 = g4.1
1/2 v^2 (A) + 31.36 = 41.16
1/2 v^2 (A) = 9.8
v^2 (A) = 19.6
V(A) = 4.427 m/s
Answered by tipur
A particle, starting from
point A in the drawing, is
projected down the
curved runway. Upon
leaving the runway at
point, the particle is
traveling straight upward
and reaches at a height of 4 m above the floor before
falling back down. Ignoring friction and air resistance,
find the speed of the particle at point A.
Answered by Anonymous
What about fo point B
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