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PhysicsKinematics2-D & 3DEasy2 minPYQ_2024

PhysicsEasysingle choice

The range of the projectile projected at an angle of $1 5^{\circ}$ with horizontal is 50 m . If the projectile is projected with same velocity at an angle of $4 5^{\circ}$ with horizontal, then its range will be :

Options:

Answer:

Solution:

Range of projectile $R = \frac{v ^{2} s i n 2 θ}{g} (∵ R \propto sin (2 θ))$ $\frac{R _{1}}{R _{2}} = \frac{s i n ( 2 θ _{1} )}{s i n ( 2 θ _{2} )} = \frac{s i n ( 2 \times 15 )}{s i n ( 2 \times 45 )} = \frac{s i n 3 0 ^{\circ}}{s i n 9 0 ^{\circ}}$ $\Rightarrow \frac{50}{R _{2}} = \frac{1}{2} \Rightarrow R_{2} = 100 m$

Stream:BITSATSubject:PhysicsTopic:KinematicsSubtopic:2-D & 3D

⏱ 2mℹ️ Source: PYQ_2024

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