Question

Figure (8-E4) shows a particle sliding on a frictionless track which terminates in a straight horizontal section. If the particle starts slipping from point A, how far away from the track will the particle hit the ground?
Figure

Answer 1

The particle will hit the ground 1 meter away from the track, a friction less which terminates in a straight horizontal section.

Explanation:

Step 1:

The particle must fall through the ending point:- h

h = 1.0 - 0.5  

h= 0.5 m

Step 2:

Initial Kinetic Energy = 0

Thus, depending on the condition, Track's termination point so it's Kinetic energy is due to the potential energy change.

$Kinetic Energy =\frac{1}{2} mv^{2}$ .

Change in potential energy = mgh .

Step 3:

When Both are equated.  

$\frac{1}{2} mv^{2} = mgh$

mv²=2mgh

$v^{2}= \frac{2mgh}{m}$

cancel 'm' on both sides of the above equation

v²=2gh

$v=\sqrt{2gh}$

$v=\sqrt{(2\times 9.8\times 0.5}$

$v=\sqrt{(19.6\times 0.5)}$

$v=\sqrt{9.8}$

v = 3.13 meter per second

Step 4:

Now the particle flies in the parabolic medium from the formula as here

so , u =0

$s=ut+\frac{1}{2} gt^{2}$

$s = 0+\frac{1}{2} gt^{2}$

$0.5 = \frac{1}{2} gt^{2}$

1 = gt²

$t^{2} = \frac{1}{g}$

$t^{2} = \frac{1}{9.8}$

t²=0.102

$t = \sqrt{0.102}$

t = 0.319  second

Distance = Speed × Time

distance = 3.13 × 0.319

distance = 0.999 ≈ 1 m

The particle thus hit the ground 1 m away from the track.

Answer 2

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Please refer the above attachment