Given: Initial velocity u0=20√2m/s; angle of projection θ=45∘
Therefore horizontal and vertical components of initial velocity are
ux=20√2cos45∘=20 m/s
and uy=20√2sin45∘=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=20m/s.
Therefore, maximum height attained by the second part will be
H=h1+h2
Using, h=ut+12at2
⇒h1=(20×1)−12×10×(1)2=15 m
a=g=10 m/s2
h2=v′2y2g=(20)22×10=20m
H=20+15=35 m.