Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 5 (Q.No. 39)
39.
A particle is dropped from the top of a tower 60 m high and another is projected upwards from the foot of the tower to meet the first particle at a height of 15.9 m. The velocity of projection of the second particle is
Discussion:
3 comments Page 1 of 1.
Shekhar pali said:
5 years ago
S = ut+.5gt^2.
Initial velocity u = 0,
44.1 = 0*t+.5*9.81*t^2.
44.1 = .5*9.81*-t^2,
44.1 = 4.905*t^2,
t^2 = 44.1/4.905,
t^2 = 8.99,
t = 2.99 = 3 sec.
Put t=3 in ek
15.9 = u*3+.5*9.81*3^2.
u*3 = 15.9+44.145,
U = 60/3,
U = 20.
Initial velocity u = 0,
44.1 = 0*t+.5*9.81*t^2.
44.1 = .5*9.81*-t^2,
44.1 = 4.905*t^2,
t^2 = 44.1/4.905,
t^2 = 8.99,
t = 2.99 = 3 sec.
Put t=3 in ek
15.9 = u*3+.5*9.81*3^2.
u*3 = 15.9+44.145,
U = 60/3,
U = 20.
Shubhamv said:
6 years ago
@Athira.
How You Got t = 3 sec.
How You Got t = 3 sec.
Athira said:
7 years ago
Distance travelled by 1st particle when they meet=44.1m.
By eqn, s= ut+0.5at^2,
We will get t=2.99sec=3 sec,
From same eqn given above,
15.9= u*3+0.5*(-9.81)*3^2,
u= 20 m/sec.
By eqn, s= ut+0.5at^2,
We will get t=2.99sec=3 sec,
From same eqn given above,
15.9= u*3+0.5*(-9.81)*3^2,
u= 20 m/sec.
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