Two particles start simultaneously from the same point and move along two straight lines, one with uniform velocity $v$ and other with a uniform acceleration $a.$ If $\alpha$ is the angle between the lines of motion of two particles then the least value of relative velocity will be at time given by

A particle moves in a plane along an elliptic path given by $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. At point $(0, b)$, the $x$-component of velocity is $u$. The $y$-component of acceleration at this point is

A hiker stands on the edge of a cliff $490\; m$ above the ground and throws a stone horizontally with an initial speed of $15 \;m/ s$. Neglecting air resistance, find the time taken by the stone to reach the ground, and the speed with which it hits the ground. (Take $g = 9.8 \;m /s^2$ ).

At a height $0.4\, m$ from the ground, the velocity of a projectile in vector form is $\vec v = \left( {6\hat i + 2\hat j} \right)\,m/{s}$. The angle of projection is ...... $^o$ $(g = 10\, m/s^2)$ 

A particle is moving along the $x-$axis with its coordinate with the time '$t$' given be $\mathrm{x}(\mathrm{t})=10+8 \mathrm{t}-3 \mathrm{t}^{2} .$ Another particle is moving the $y-$axis with its coordinate as a function of time given by $\mathrm{y}(\mathrm{t})=5-8 \mathrm{t}^{3} .$ At $\mathrm{t}=1\; \mathrm{s},$ the speed of the second particle as measured in the frame of the first particle is given as $\sqrt{\mathrm{v}} .$ Then $\mathrm{v}$ (in $\mathrm{m} / \mathrm{s})$ is

A particle moves in space along the path $z = ax^3 + by^2$ in such a way that $\frac{dx}{dt} = c = \frac{dy}{dt}.$ Where $a, b$ and $c$ are contants. The acceleration of the  particle is