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Applied Mechanics - Engineering

Q1:

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

A 16 m/sec

B 18 m/sec

C 20 m/sec

D 22 m/sec

ANS:C - 20 m/sec

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.