Question

A particle of mass 2m is projected at an angle of 30 degree with the horizontal with a velocity of 40 m/s after 1s explosion takes place and the particle is broken into two equal pieces .As a result of explosion one part comes to rest.The maximum height from the ground attained by the other parts is

Answer 1

Given that,

Mass of particle = 2 m

Angle = 30°

Velocity = 40 m/s

Time = 1 sec

We need to calculate the maximum velocity

Using formula of vertical component

$v_{y}=v\sin\theta$

$v_{y}=40\sin30$

$v_{y}=20\ m/s$

We need to calculate the maximum height from the ground attained by the other parts

Using formula of maximum height

$h=ut-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=20\times1-\dfrac{1}{2}\times10\times1^2$

$h=15\ m$

Hence, The maximum height from the ground attained by the other part is 15 m.