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Ludmilka [50]
3 years ago
8

A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight upwar

d at the same instant that the first ball is dropped. The initial speed of the second ball is exactly the same as that with which the first ball eventually hits the ground. In the absence of air resistance, the motions of the balls are just the reverse of each other. Determine how far below the top of the cliff the balls cross paths
Physics
1 answer:
sesenic [268]3 years ago
4 0

Answer:

6.0 m below the top of the cliff

Explanation:

We can find the velocity at which the ball dropped from the cliff reaches the ground by using the SUVAT equation

v^2-u^2 = 2gd

where

u = 0 (it starts from rest)

g = 9.8 m/s^2 (acceleration of gravity, we assume downward as positive direction)

h = 24 m is the distance covered

Solving for h,

v=\sqrt{2gh}=\sqrt{2(9.8)(24)}=21.7 m/s

So the ball thrown upward is launched with this initial velocity:

u = 21.7 m/s

From now on, we take instead upward as positive direction.

The vertical position of the ball dropped from the cliff at time t is

y_1 = h - \frac{1}{2}gt^2

While the vertical position of the ball thrown upward is

y_2 = ut - \frac{1}{2}gt^2

The two balls meet when

y_1 = y_2\\h-\frac{1}{2}gt^2 = ut - \frac{1}{2}gt^2 \\h = ut \rightarrow t = \frac{h}{u}=\frac{24}{21.7}=1.11 s

So the two balls meet after 1.11 s, when the position of the ball dropped from the cliff is

y_1 = h -\frac{1}{2}gt^2 = 24-\frac{1}{2}(9.8)(1.11)^2=18.0 m

So the distance below the top of the cliff is

d=24.0 - 18.0 = 6.0 m

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Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

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Therefore, the required work is 29.596 kJ.

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