AP EAMCET · PHYSICS · Motion In Two Dimensions
A projectile is launched from the ground, such that it hits a target on the ground which is \(90 \mathrm{~m}\) away. The minimum velocity of projectile to hit the target is (acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )
- A 10. \(\mathrm{ms}^{-1}\)
- B \(16 \mathrm{~ms}^{-1}\)
- C \(60 \mathrm{~ms}^{-1}\)
- D \(30 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(D) \(30 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Since, launched projectile hit the target on the ground \(90 \mathrm{~m}\) away hence, range, \(R=90 \mathrm{~m}\). For minimum velocity \(\left(u_{\min }\right)\) of projectile, range should be maximum. i.e. \(\theta=45^{\circ}\) \(R=\frac{u^2 \sin 2 \theta}{g}\)…
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