8.
A bullet travelling horizontally looses 1/20th
of its velocity while piercing a wooden plank.
Then the number of such planks required to
stop the bullet is
1) 6 2) 9 3) 11
4) 13
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8.
A bullet travelling horizontally looses 1/20th
of its velocity while piercing a wooden plank.
Then the number of such planks required to
stop the bullet is
1) 6 2) 9 3) 11
4) 13
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Answer:
3) 11
Explanation:
If the bullet looses 1/20th of its velocity then
V= 19/20u
Since, v² = u² + 2as
(19/20u)² = u² - 2as
2as = u² - (19/20u)² ................(1)
Now, initial velocity = 0
Therefore 0² = u² - 2a (ns)
u² = 2as (n)
u² = n [ u² - 19²/20² u²]
u² = nu² [ 20²-19²/20²]
20² = n [ 20² - 19² ]
20² = n [ (20+19)(20-19) ]
n = 400/39
n ~ 11 planks