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Motion in A Plane question

2000 · Q147
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Motion in A Plane question

2000 · Q147

NEETPhysicsMotion in A PlaneMCQ+4 / −1
Two projectiles of same mass and with same velocity are thrown at an angle 60o and 30o with the horizontal, then which will remain same
  1. A
    time of flight
  2. B
    range of projectile
  3. C
    maximum height acquired
  4. D
    all of them.
View written solutionFree

Correct answer: B

Given, u1 = u2 = u, θ1\theta _1θ1​ = 60º, θ2\theta _2θ2​ = 30º

In 1st case, we know that range
R1=u2sin⁡2(60∘)g=u2sin⁡120∘g{R_1} = {{{u^2}\sin 2\left( {60^\circ } \right)} \over g} = {{{u^2}\sin 120^\circ } \over g}R1​=gu2sin2(60∘)​=gu2sin120∘​
=u2sin⁡(90∘+30∘)g = {{{u^2}\sin \left( {90^\circ + 30^\circ } \right)} \over g}=gu2sin(90∘+30∘)​
=u2(cos⁡30∘)g=u232g = {{{u^2}\left( {\cos 30^\circ } \right)} \over g} = {{ {u^2}\sqrt 3} \over {2g}}=gu2(cos30∘)​=2gu23​​

In 2nd case, when θ2\theta _2θ2​ = 30° , then
R2=u2sin⁡60∘g=u232g{R_2} = {{{u^2}\sin 60^\circ } \over g} = {{{u^2}\sqrt 3 } \over {2g}}R2​=gu2sin60∘​=2gu23​​

⇒R1=R2 \Rightarrow {R_1} = {R_2}⇒R1​=R2​

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