Q124 / 160 · AP EAMCET 2019 (23 Apr Shift 1) · Physics
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\(50 \mathrm{~m}\)
- Correct answerD\(60 \mathrm{~m}\)
Solution
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 \(\begin{array}{ll} \Rightarrow & 20=\frac{u}{\sqrt{2}} t-5 t^2 \\ \Rightarrow \quad & t=\sqrt{2} \mathrm{~s} \text { and } u=30 \mathrm{~m} / \mathrm{s} \\ \text {Hence, } R= & \frac{u^2 \sin 2 \theta}{g}=\frac{30 \times 30 \times \sin 90^{\circ}}{10}=90 \mathrm{~m} \end{array}\) \(\therefore\) Distance between pole and the point at which body strike on the ground, \(\Rightarrow \quad s=(90-30)=60 \mathrm{~m}\) So, the correct option is (d).
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 \(\begin{array}{ll} \Rightarrow & 20=\frac{u}{\sqrt{2}} t-5 t^2 \\ \Rightarrow \quad & t=\sqrt{2} \mathrm{~s} \text { and } u=30 \mathrm{~m} / \mathrm{s} \\ \text {Hence, } R= & \frac{u^2 \sin 2 \theta}{g}=\frac{30 \times 30 \times \sin 90^{\circ}}{10}=90 \mathrm{~m} \end{array}\) \(\therefore\) Distance between pole and the point at which body strike on the ground, \(\Rightarrow \quad s=(90-30)=60 \mathrm{~m}\) So, the correct option is (d).
Hence, the correct answer is option: 'D'.