v = u + at

u = 3i + 4j + (0.4 i + 0.3 j) × 10

= 3i + 4j + 4i + 3i

u = 7i + 7j

$|\overrightarrow{\mathrm{u}}|=\sqrt{7^{2}+7^{2}}$

$=7 \sqrt{2}$

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

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

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

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

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

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

2 × 0.25 = c

5 = c

$2 \sqrt{\mathrm{v}}=-\mathrm{kt}+5$

0 = –kt + 5

kt = 5

$\mathrm{t}=\frac{5 \mathrm{sec}}{2.5}=2 \mathrm{sec}$

Time to reach highest point $=\mathrm{t}=\frac{\mathrm{u}}{\mathrm{g}}$

time to reach ground = nt

$\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$

$-\mathrm{H}=\mathrm{u}(\mathrm{nt})-\frac{1}{2} \mathrm{g}(\mathrm{nt})^{2}$

$\Rightarrow 2 \mathrm{gH}=\mathrm{nu}^{2}(\mathrm{n}-2)$

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

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)