Question

44. A body is projected vertically upward. Show that th
(i) distance covered by it in the first 'n' seconds of
ascent is equal to the distance covered by it in the
first 'n' seconds of its descent.
(i) the distance covered by it in the last 'n' seconds
of ascent is equal to the distance covered by it in
the first 'n' seconds of its descent.​

Answer 1

Answer:

Part a)

Distance moved by it during first n second of upward trip and last n second of downward trip will be same and given as

$d = v(n) - \frac{1}{2}gn^2$

Part b)

the distance covered by it in the last 'n' seconds  of ascent is equal to the distance covered by it in  the first 'n' seconds of its descent.​

$d = \frac{1}{2}gn^2$

Explanation:

Part a)

Let the object is projected up with speed v

so the total time for which it will move upwards is given as

$t = \frac{v}{g}$

now we have know that the distance moved by it in first n seconds is given as

$d = v(n) - \frac{1}{2}gn^2$

now during its return journey the distance covered by it in last n second is given as

$d = \frac{1}{2}g(\frac{v}{g})^2 - \frac{1}{2}g(\frac{v}{g} - n)^2$

$d = \frac{1}{2}g(\frac{2v}{g} - n)(n)$

$d = vn - \frac{1}{2}gn^2$

So the distance moved in first n second in upward trip is same as distance moved in last n second of return trip

Part b)

Now similarly the distance moved by it in last n second of upward trip is

$d =\frac{v^2}{2g} - v(\frac{v}{g} - n) + \frac{1}{2}g(\frac{v}{g} - n)^2$

$d =\frac{v^2}{2g} - \frac{v^2}{g} + vn + \frac{1}{2}g(\frac{v}{g} - n)^2$

$d = -\frac{v^2}{2g} + vn + \frac{v^2}{2g} + \frac{n^2g}{2} - vn$

$d = \frac{1}{2}gn^2$

Now during its return trip the distance moved in first n second is

$d = \frac{1}{2}gn^2$

#Learn

Topic : Kinematics

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