TS EAMCET · Physics · Motion In Two Dimensions
A particle aimed at a target, projected with an angle \(15^{\circ}\) with the horizontal is short of the target by \(10 \mathrm{~m}\). If projected with an angle of \(45^{\circ}\) is away from the target by \(10 \mathrm{~m}\), then the angle of projection to hit the target is
- A \(\frac{1}{2} \sin ^{-1}\left(\frac{1}{4}\right)\)
- B \(\frac{1}{2} \sin ^{-1}\left(\frac{3}{4}\right)\)
- C \(\frac{1}{2} \sin ^{-1}\left(\frac{10}{4}\right)\)
- D \(\frac{1}{2} \sin ^{-1}\left(\frac{20}{4}\right)\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2} \sin ^{-1}\left(\frac{3}{4}\right)\)
Step-by-step Solution
Detailed explanation
Let \(d\) be the distance of target. The horizontal range of projectile is \(R=\frac{u^2 \sin 2 \theta}{g}\) When \(\theta=15^{\circ}\), \(R_1=\frac{u^2 \sin \left(2 \times 15^{\circ}\right)}{g}=\frac{u^2}{2 g}\) When \(\theta=45^{\circ}\),…
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