BITSAT 2015 Paper

Given : Initial velocity µ0 = 20 √2 m/s ;
angle of projection θ = 45°
Therefore horizontal and vertical components of initial velocity are
ux = 20√2 cos 450 = 20 m/s
and uy = 20√2 sin 45° = 20 m/s
After 1s, horizontal component remains unchanged while the vertical component becomes
vy = uy - gt
Due to explosion, one part comes to rest.
Hence, from the conservation of linear momentum, vertical component of second part will become
v′y = 20 m/s
Therefore, maximum height attained by the second part will be
H = h1+h2
Using, h = ut +

1

2

at2
⇒ h1 = (20 ×1) -

1

2

×10×(1)2 = 15 m
a = g = 10 m/s2
h2 =

v′y2

2g

=

(20)2

2×10

= 20 m
H = 20 + 15 = 35 m