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A stone thrown vertically upward with initial velocity u reach a hight h before coming down.show that the time taken to go up is same as time taken to go down
Asked by Pardeshivinay7 | 26 Apr, 2019, 07:26: AM
We need to use the equations of motion,

v = u + a×t   .....................(1)
v2 = u2 + 2×a×S .................(2)

where v is final velocity after time t, u is initial velocity, a is acceleration and S is distance travelled in time t.

Let u be the initial velocity of stone.
when the stone reaches the maximum height after time t, its final velocity is zero.
When the stone travelling vertically upwards, it is experiencing the retardation due to gravity.

hence we write the above equation of motion (1) as, 0 = u - g×t or t = u/g .............(3)

eqn.(2) will be written as,   0 = u2 - 2×g×h,  or  ...................(4)

by substituting u from eqn.(4) in eqn.(3), we have t............(5)
After reaching maximum height, the stone descends with zero initial velocity,
accelerated downwards due to gravity and reaches the ground after time t'.

Eqns.(1) and (2) are written as follows for downward travel

v = 0 + gt'  or  t' = v/g ..........................(6)

v2 = 0 + 2gh or.....................(7)

from eqn.(4) and eqn.(7), we can conclude that initial projection velocity and
final velocity when the stone reaches the ground are equal

by substituting for v from eqn.(7) in eqn.(6), we get , ................(8)
from eqn.(5) and eqn.(8),  time of ascending the height h is equal to time of descending from same height h.
Answered by Shiwani Sawant | 26 Apr, 2019, 01:06: PM
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