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Previous Years AIEEE/JEE Mains Questions
(1) 7 units (2) 8.5 units (3) 10 units (4) $7 \sqrt{2}$ units [AIEEE-2009]
v = u + at u = 3i + 4j + (0.4 i + 0.3 j) × 10 = 3i + 4j + 4i + 3i u = 7i + 7j $|\overrightarrow{\mathrm{u}}|=\sqrt{7^{2}+7^{2}}$ $=7 \sqrt{2}$
(1) 4 s (2) 8 s (3) 1 s (4) 2 s [AIEEE-2011]
$\overrightarrow{\mathrm{a}}=\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}$ $\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=-2.5 \sqrt{\mathrm{v}}$ $\int \mathrm{d} \mathrm{v} \times \mathrm{v}^{-1 / 2}=\int-2.5 \mathrm{dt}$ $\frac{\mathrm{v}^{\frac{1}{2}}}{\frac{1}{2}}=-2.5 \mathrm{t}$ $2 \sqrt{\mathrm{v}}=-2.5 \mathrm{t}+\mathrm{c}$ $2 \sqrt{6.25}=-2.5 \mathrm{t}+\mathrm{c}$ 2 × 0.25 = c 5 = c $2 \sqrt{\mathrm{v}}=-\mathrm{kt}+5$ 0 = –kt + 5 kt = 5 $\mathrm{t}=\frac{5 \mathrm{sec}}{2.5}=2 \mathrm{sec}$
(1) $2 \mathrm{g} \mathrm{H}=\mathrm{nu}^{2}(\mathrm{n}-2)$ (2) $\mathrm{g} \mathrm{H}=(\mathrm{n}-2) \mathrm{u}^{2}$ (3) $2 \mathrm{g} \mathrm{H}=\mathrm{n}^{2} \mathrm{u}^{2}$ (4) $\mathrm{g} \mathrm{H}=(\mathrm{n}-2)^{2} \mathrm{u}^{2}$ [JEE-MAIN-2014]
Time to reach highest point $=\mathrm{t}=\frac{\mathrm{u}}{\mathrm{g}}$ time to reach ground = nt $\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$ $-\mathrm{H}=\mathrm{u}(\mathrm{nt})-\frac{1}{2} \mathrm{g}(\mathrm{nt})^{2}$ $\Rightarrow 2 \mathrm{gH}=\mathrm{nu}^{2}(\mathrm{n}-2)$
[JEE Mains – 2017]
Velocity at any time t is given by v = u + at v = v0 + (–g)t $\mathrm{v}=\mathrm{v}_{0}-\mathrm{gt}$ $\Rightarrow$ straight line with negative slope
[JEE Mains – 2018]
In this question option (2) and (4) are the corresponding position – time graph and velocity –position graph of option (3) and its distance – time graph is given as Hence incorrect graph is option (1)
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