TS EAMCET · Physics · Motion In Two Dimensions
A cannon placed on a cliff at a height of 375 m fires a cannon ball with a velocity of \(100 \mathrm{~ms}^{-1}\) at an angle of \(30^{\circ}\) above the horizontal. The horizontal distance between the cannon and the target is (Acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )
- A \(750 \sqrt{3} \mathrm{~m}\)
- B \(500 \sqrt{3} \mathrm{~m}\)
- C \(250 \sqrt{3} \mathrm{~m}\)
- D 750 m
Answer & Solution
Correct Answer
(A) \(750 \sqrt{3} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
Here vertical component of velocity \(u=-100 \sin 30^{\circ}\) \(=-50 \mathrm{~m} / \mathrm{s}\) Let \(t\) be the time taken to reach…
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