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A ball is thrown vertically downwards at speed vo from height h. Draw velocity vs. time & acceleration vs. time graphs. In terms of only the givens vo and h, derive expressions for the final speed of the ball and the elapsed time of flight.

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Answer:

Explanation:

let the ball is thrown vertically downwards with velocity u.

So, initial velocity, = - u (downwards)

acceleration = - g (downwards)

let the velocity is v after time t.

use first equation of motion

v = u + at

- v = - u - gt

v = u + gt

So, it is a straight line having slope g and y intersept is u.

The graph I shows the velocity - time graph.

Now the value of acceleration remains constant and it is equal to - 9.8 m/s^2.

So, acceleration time graph is a starigh line parallel to time axis having slope zero.

the graph II shows the acceleration - time graph.

Use III equation of motion to find the final speed in terms height.

[tex]v^{2}=u^{2}+2gh[/tex]

And the time is

v = u + gt

[tex]t=\frac{v-u}{g}[/tex]

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