Question

d. A stone thrown vertically upwards with initial velocity u reaches a height 'h' before coming down. Show that the
time taken to go up is same as the time taken to come down.​

Answer 1

When stone is thrown vertically upwards with initial velocity , u ,

final velocity at maximum height, v = 0

Apply kinematic equation of motion

$\bold {v = u + at}$

$\sf {\implies t = \frac {v - u}{a} = \frac {0 - u}{-g }}$

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

$\sf{\implies h = u(\frac {u}{g} ) − \frac {1}{2}g( \frac {u}{g} )^{2}}$

$\sf { \implies u = \sqrt{2gh}}$

Time ,

$\sf { t = \frac {u}{g} = \frac{\sqrt{2gh}}{g}}$

$\sf {\implies t = \sqrt {\frac{2h}{g}} \: \: \: (1)}$

When stone is dropped from height h ,

Apply kinetic equation of motion

$\bold { v^{2} - u^{2} = 2gh}$

$\sf{\implies v = \sqrt {2gh}}$

$\bold{v = u + at }$

$\sf {\implies t = \frac{v - u}{a} = \frac{\sqrt{2gh} - 0}{g} = \sqrt {\frac{2h}{g}} \: \: \: (2)}$

From (1 ) and (2) it is proved that the time taken to go up is same as the time taken to come down.

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Answer 2

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