Question

A projectile is fired at speed v0 at an angle ‘θ’ such that range of the projectile is R. At top-most point if it breaks into two identical parts such that one of them comes to rest, then horizontal range of other particle measured from the point of projection, is

Answer 1

Answer:

for this question the answer should be 7r/4

Answer 2

Answer:

The horizontal range of particle measured from point of projection is

$R_{0} = R( \frac{1 + 8cos {}^{2}θ }{2} )$

Explanation:

Given that,projectile is fired at a speed v0 at an angle ‘θ' such that range of the projectile is R and At top-most point if it breaks into two identical parts such that one of them comes to rest.

Let the mass of projectile=m

=>Mass of two identical parts=m/2

Using conservation of momentum at maximum height

$mu \cosθ = \frac{m}{2} (0) + \frac{m}{2} v \\ v = 2ucosθ$

Therefore velocity of other particle is v=2ucosθ

Now the Range of this part is

$R {}^{1} = \frac{v {}^{2} \sin2θ}{g} = \frac{4u {}^{2}cos {}^{2} θ \sin2θ}{g} = 4cos {}^{2} θ( \frac{u {}^{2}sin2θ }{g} ) = 4cos {}^{2} θ(R)$

To find Range measured from initial point of projection,we add R/2 to the obtained range.

Therefore original range is

$R_{0} = \frac{R}{2} + R(4cos {}^{2} θ ) \\ = R( \frac{1 + 8cos {}^{2}θ }{2} )$

Hence the horizontal range of particle measured from point of projection is obtained.

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