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*Simulator*

**Previous Years AIEEE/JEE Mains Questions**

(1) 7 units

(2) 8.5 units

(3) 10 units

(4) $7 \sqrt{2}$ units

**[AIEEE-2009]**

**Sol.**(4)

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]**

**Sol.**(4)

$\overrightarrow{\mathrm{a}}=\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}$

$\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=-2.5 \sqrt{\mathrm{v}}$

$\int \mathrm{d} \mathrm{v} \times \mathrm{v}^{-1 / 2}=\int-2.5 \mathrm{dt}$

$\frac{\mathrm{v}^{\frac{1}{2}}}{\frac{1}{2}}=-2.5 \mathrm{t}$

$2 \sqrt{\mathrm{v}}=-2.5 \mathrm{t}+\mathrm{c}$

$2 \sqrt{6.25}=-2.5 \mathrm{t}+\mathrm{c}$

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]**

**Sol.**(1)

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]**

**Sol.**(1)

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]**

**Sol.**(1)

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)

Nice