85. A particle initially at origin starts moving in $X Y$ - plane has velocity component $\mathbf{v}=(6+2 t) \hat{\mathbf{i}}+(4+2 \sqrt{3 t}) \hat{\mathbf{j}} \mathrm{m} / \mathrm{s}$. Acceleration of the particle in $\mathrm{m} / \mathrm{s}^2$ is $[x, y$ are measured in meters, $t$ in seconds, respectively

A bullet is fired at time $t=0$ with velocity $20 \mathrm{~m} / \mathrm{s}$ and at an initial angle of $30^{\circ}$ with the horizontal. The angle between the displacement vector and the horizontal after time 0.1 s is (assume $g=10 \mathrm{~m} / \mathrm{s}^2$ ).

A man walking along a straight line with a velocity 6 $\mathrm{km} / \mathrm{h}$ encounters rain falling vertically down with a velocity $6 \sqrt{3} \mathrm{~km} / \mathrm{h}$. At what angle the man should hold his umbrella, so that he can protect himself from rain

A projectile is given an initial velocity of $(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$ where $\hat{\mathbf{i}}$ is along the ground and $\hat{\mathbf{j}}$ is along the vertical. Assuming $g=10 \mathrm{~m} / \mathrm{s}^2$, if the equation of its trajectory can be written as $\frac{1}{9}\left[\beta x+\gamma x^2\right]$. Then the value of $\gamma$ is