Question

A particle starting from point A in the drawing at a height ho = 2.4 m, is protected down the curved runway. upon leaving the runway at point B the particle is traveling straight upward and reaches a height hf = 4.3 m above the floor before falling back down. ignoring friction and air resistance, find the speed of the particle at point A.​

Answer 1

Answer:

Total energy at A total energy at B

mg*3.2  + KE1 mg*4.1 + 0 solve for KE1, then solve v1 from

v1 = sqrt (KE1/mg)

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///62 (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

1/2v^2(A) + 31.36 = 41.16

1/2 V^2 (A) = 9.8

v^2(A) = 19.6

V(A) = 4.427 m/s

Explanation:

Answer 2

Given: Initial position's height of particle ho= 2.4m, final position's height of particle hf= 4.3m

To find: speed of the particle at point A

Solution: To find the speed of the particle at position A we will first write an equation conserving the total energy of the particle at both the positions that are

PE(A) + KE(A) = PE(B) + KE(B)

here PE is potential energy and KE is kinetic energy

mgho + 1/2mvo^2 = mghf+ 1/2mvf^2

here ho and vo is the height and velocity of a particle at position A and hf and vf is the height and velocity of a particle at position B

m will be the same in both the positions

9.8×2.4 + 1/2vo^2 = 9.8×4.3 + 1/2(0)^2

vo^2 = 37.24m/s

vo = 6.1 m/s

Therefore, the speed of the particle at point A will be 6.1 m/s.