AP EAMCET · PHYSICS · Motion In Two Dimensions
A body is projected at an angle of \(45^{\circ}\) from a point on the ground at a distance of \(30 \mathrm{~m}\) from the foot of a vertical pole of height \(20 \mathrm{~m}\). The body just crosses the top of the pole and strikes the ground at a distance \(s\) from the foot of the pole on the other side of the pole. Then, \(s\)
- A \(20 \mathrm{~m}\)
- B \(30 \mathrm{~m}\)
- C \(50 \mathrm{~m}\)
- D \(60 \mathrm{~m}\)
Answer & Solution
Correct Answer
(D) \(60 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
According to the question, projection of a body shown in the figure below, Now, from the above figure, let \(t\) be the time of crossing of pole, then, \(30=t u \cos 45^{\circ}\) ...(i) and \(20=u \sin 45^{\circ} t-\frac{10}{2} t^2\) ...(ii) Hence, \(30 \sqrt{2}=u t\) and…
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