Q. No.
Consider the two situations shown in the figures. In figure (A), the particle \(P\) is given a velocity \(u\) up a smooth horizontal incline and it reaches a maximum vertical height \(h_A\). In figure (B), the same particle \(P\) is projected with a velocity \(u\) at an angle \(\theta\) (parallel to the previous incline) and reaches a maximum height \(h_B\).
Then,
1.
\(h_A=h_B~\text{sin}\theta\)
2.
\(h_A~\text{sin}\theta=h_B\)
3.
\(h_A~\text{sin}^2\theta=h_B\)
4.
\(\dfrac{h_A}{\text{sin}^2\theta}=h_B\)
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It can be concluded that:
| 1. | \(v_1>v_2\) |
| 2. | \(v_1<v_2\) |
| 3. | \(v_1=v_2\) |
| 4. | Any of the above can be true depending on the angle of projection |
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| 1. | \(u_1=\dfrac{\sqrt3}{2}u_2\) | 2. | \(u_1=\dfrac{1}{2}u_2\) |
| 3. | \(u_1=\dfrac{1}{\sqrt2}u_2\) | 4. | \(u_1=\dfrac{1}{\sqrt3}u_2\) |
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1. \(v=4u\)
2. \(v=2u\)
3. \(v=u\)
4. \(v<u\)
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1. \(a\)
2. \(b\)
3. \(c\)
4. \(d\)
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1. \(1\) min
2. \(6\) min
3. \(0.5\) min
4. \(\dfrac {1}{6}\) min
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| 1. | \(2\) km | 2. | \(1\) km |
| 3. | \(\dfrac12\) km | 4. | \(\dfrac14\) km |
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| 1. | zero | 2. | \(\dfrac{2u^2}{R}\) |
| 3. | \(\dfrac{u^2}{\sqrt2R}\) | 4. | \(\dfrac{\sqrt2u^2}{R}\) |
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| 1. | \(10\) min | 2. | \(5\sqrt3\) min |
| 3. | \(20\) min | 4. | \(\dfrac{10}{\sqrt3}\) min |
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1. \(2.4\) km
2. \(1.6\sqrt{3}\) km
3. \(1.6\sqrt2\) km
4. \(3.2\) km
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