A particle of mass 2m is projected at an angle of 45º with horizontal with a velocity of 
m/s. After 1 s explosion takes place and the particle is broken into two equal pieces. As a result of explosion one part comes to rest. Find the maximum height attained from the ground by the other part. Take g = 10 m/s 2 .
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(35)
u y = usin 45º = 20 u x = u cos 45º = 20 m/sec.
After 1 sec.
V y = u y – gt = 20 – 10 = 10 V x = u x = 20 m/sec.
When explosion takes place one particle goes with double velocity and other goes to rest.
In x direction 2 m V x = m × 0 + m V x ' ∴ V x ' = 2V x
In y direction 2 mV y = m × 0 + m V y ' ∴ V y ' = 2V y
h 1 = u y t – 
gt 2 = 20 × 1 – 5 × 1 = 15 m.
h 2 = 
= 20 ∴ h 1 + h 2 = 15 + 20 = 35 m.
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