If a particle is thrown vertically upwards, then its velocity so that it covers same distance in 5th and 6th second would be
If a particle is thrown vertically upwards, then its velocity so that it covers same distance in 5th and 6th second would be
The correct option is C 49 m/s
Let u be the
initial velocity of the ball. Distance travelled in the nth sec is
given by,
Sn=u+12×a×(2n–1)
Distance covered in
5th second:
(put a = -9.8 and n
= 5)
S5=u–44.1
Distance covered in
the 6th second:
(put n =6, a=-9.8 in
equation):
S6=u–53.9
Now, as the body is
projected upwards, the distance travelled in two different time
intervals will be
same only when at
one time it is moving upwards and the other time it is coming
downwards (as there is symmetry in time and height). So, during 5th
second, the object was moving upwards, and during the 6th second, the
object is coming downwards. Therefore, the distance S6 will be
negative.
S6=−(u–53.9)
Since they are
equal, equating S5 and S6 :
u−44.1=−u+53.9
2u=98
u=49m/s
Let u be the initial velocity of the ball. Distance travelled in the nth sec is given by,
Sn=u+12×a×(2n–1)
Distance covered in 5th second:
(put a = -9.8 and n = 5)
S5=u–44.1
Distance covered in the 6th second:
(put n =6, a=-9.8 in equation):
S6=u–53.9
Now, as the body is projected upwards, the distance travelled in two different time intervals will be
same only when at one time it is moving upwards and the other time it is coming downwards (as there is symmetry in time and height). So, during 5th second, the object was moving upwards, and during the 6th second, the object is coming downwards. Therefore, the distance S6 will be
negative.
S6=−(u–53.9)
Since they are equal, equating S5 and S6 :
u−44.1=−u+53.9
2u=98
u=49m/s
