Question

A particle of mass 2m is projected at an angle 45 with the horizontal with a velocity of 20root2 m/s.

Answer 1

The maximum height attained by the other part is 35 m.

Explanation:

Given that,

Mass = 2 m

Angle = 45

Velocity = 20√2 m/s

Suppose 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 by the other part

Let velocity of second part is v.

We need to calculate the velocity after explosion

Using equation of motion

]$v_{y}u+gt$

Put the value into the formula

$v_{y}=20\sqrt{2}\sin45-10\times1$

$v_{y}=10j\ m/s$

Using conservation of motion

$mv=mu+mv'$

Put the value into the formula

$2m\times(20\sqrt{2}\cos45 i+10 j)=0+mv$

$v=(40i+20j)\ m/s$

We need to calculate the height

Using equation of motion

$v^2=u^2+2as$

$0=400+2\times(-10)\times s$

$s=20\ m$

We need to calculate the initial height

Using equation of motion

$h_{i}=ut+\dfrac{1}{2}gt^2$

$h_{i}=20\times1-\dfrac{1}{2}\times10\times1$

$h_{i}=15\ m$

We need to calculate the maximum height attained by the other part

Using formula of height

$h=h_{i}+\dfrac{u^2}{2g}$

Put the value into the formula

$h=15+\dfrac{20^2}{2\times10}$

$h=35\ m$

Hence, The maximum height attained by the other part is 35 m.

Learn more :

Topic : Projectile motion

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