KCET · Physics · Motion In Two Dimensions
Three projectiles \( A, B \) and \( C \) are projected at an angle of \( 30^{\circ}, 45^{\circ}, 60^{\circ} \) respectively. If
\( R_{A}, R_{B} \) and \( R_{C} \) are ranges of \( A, B \) and \( C \) respectively, then (velocity of projection is same for \( A \),
\( B \& C) \)
- A \( R_{A}=R_{B}=R_{C} \)
- B \( R_{A}=R_{C}>R_{B} \)
- C \( R_{A} < R_{B} < R_{C} \)
- D \( R_{A}=R_{C} < R_{B} \)
Answer & Solution
Correct Answer
(D) \( R_{A}=R_{C} < R_{B} \)
Step-by-step Solution
Detailed explanation
Given, angle of projection of \(A=30^{\circ} ;\) angle of projection of \(B=45^{\circ} ;\) angle of projection of \(C=60^{\circ} .\) Range of \(A=R A ;\)
rangeof \(B=R_{B}\) irange of \(C=R_{C}\)
Now, Horizontal range
\(=\frac{(\text { Initial velocity })^{2} \times \sin (2 \times \text { Launch angle })}{2 \times \text { Acceleration due to gravity }}\)
\(\Rightarrow R=\frac{v_{0}^{2} \sin 2 \theta}{g}\)
\(\Rightarrow R \propto \sin 2 \theta\)
\(\therefore R_{A} \propto \sin \left(2 \times 30^{\circ}\right)=\sin \left(60^{\circ}\right)\)
\(\Rightarrow R_{A}=\frac{\sqrt{3}}{2}\)
\(R_{B} \propto \sin \left(2 \times 45^{\circ}\right)=\sin \left(90^{\circ}\right)\)
\(\Rightarrow R_{B}=1\)
\(R_{C} \propto \sin \left(2 \times 60^{\circ}\right)=\sin \left(120^{\circ}\right)\)
\(\Rightarrow R_{C}=\frac{\sqrt{3}}{2}\)
Thus, \(R_{A}=R_{C} < R_{B}\)
rangeof \(B=R_{B}\) irange of \(C=R_{C}\)
Now, Horizontal range
\(=\frac{(\text { Initial velocity })^{2} \times \sin (2 \times \text { Launch angle })}{2 \times \text { Acceleration due to gravity }}\)
\(\Rightarrow R=\frac{v_{0}^{2} \sin 2 \theta}{g}\)
\(\Rightarrow R \propto \sin 2 \theta\)
\(\therefore R_{A} \propto \sin \left(2 \times 30^{\circ}\right)=\sin \left(60^{\circ}\right)\)
\(\Rightarrow R_{A}=\frac{\sqrt{3}}{2}\)
\(R_{B} \propto \sin \left(2 \times 45^{\circ}\right)=\sin \left(90^{\circ}\right)\)
\(\Rightarrow R_{B}=1\)
\(R_{C} \propto \sin \left(2 \times 60^{\circ}\right)=\sin \left(120^{\circ}\right)\)
\(\Rightarrow R_{C}=\frac{\sqrt{3}}{2}\)
Thus, \(R_{A}=R_{C} < R_{B}\)
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