Velocity, Momentum & Energy in Flight

A batsman hits a ball with a velocity 'v', making an angle of $60^\circ$ with the vertical. After some time direction of velocity is making an angle of $60^\circ$ with the horizontal. The speed of the ball at this instant is

$\left[\cos(60^\circ) = \frac{1}{2},\ \cos(30^\circ) = \frac{\sqrt{3}}{2}\right]$

If $x$ and $y$ coordinates of a projectile as a function of time $(t)$ are given as $24t$ and $43.6t - 4.9t^{2}$, respectively, then the angle (in degrees) made by the projectile with horizontal when $t = 2\text{ s}$ is ______.

A projectile is thrown upward at an angle $60^\circ$ with the horizontal. The speed of the projectile is $20\text{ m/s}$ when its direction of motion is $45^\circ$ with the horizontal. The initial speed of the projectile is $\underline{\qquad}\text{ m/s}$.

A particle is projected from ground with speed u and at an angle $\theta$ with horizontal. If at maximum height from ground, the speed of a particle is half times its initial velocity of projection then its maximum height attained is $(\cos 60^\circ = 0.5,\ \sin 60^\circ = \sqrt{3}/2)$ (g = acceleration due to gravity)

A ball P is projected at an angle of $60^\circ$ with the vertical with certain initial speed. Another ball Q of the same mass as that of ball P is projected vertically upwards with the same initial speed as that of P. At the highest point, the ratio of potential energy of ball P to that of ball Q is

$(\sin 30^\circ = 0.5)$

A ball of mass 100 g is projected with velocity $20\text{ m/s}$ at $60^\circ$ with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is

An object of mass ' m ' is projected from origin in a vertical xy plane at an angle $45^\circ$ with the $x\text{-}$ axis with an initial velocity $v_0$. The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be [ g is acceleration due to gravity]

For a body projected at an angle with the horizontal from the ground, choose the correct statement.

A ball is projected from point A with velocity $20\text{ m s}^{-1}$ at an angle $60^\circ$ to the horizontal direction. At the highest point $\text{B}$ of the path (as shown in figure), the velocity $v\text{ ms}^{-1}$ of the ball will be:

A projectile is projected at $30^\circ$ from horizontal with initial velocity $40\text{ ms}^{-1}$. The velocity of the projectile at $t = 2\text{ s}$ from the start will be: (Given $g = 10\text{ m/s}^{2}$)

A ball is projected with a velocity, 10 ms${}^{-1}$, at an angle of 60$^\circ$ with the vertical direction. Its speed at the highest point of its trajectory will be

A projectile is launched at an angle '$\alpha$' with the horizontal with a velocity 20 ms${}^{-1}$. After 10 s, its inclination with horizontal is '$\beta$'. The value of tan$\beta$ will be: (g = 10 ms${}^{-2}$).

A body is projected from the ground at an angle of 45 $^\circ$ with the horizontal. Its velocity after 2s is 20 ms${}^{-1}$. The maximum height reached by the body during its motion is __________ m. (use g = 10 ms${}^{-2}$)

A body of mass $10\text{ kg}$ is projected at an angle of $45^\circ$ with the horizontal. The trajectory of the body is observed to pass through a point $(20, 10)$. If $T$ is the time of flight, then its momentum vector, at time $t = \frac{T}{\sqrt{2}}$, is _____________.

[Take $g = 10\text{ m/s}^{2}$ ]

The velocity of a projectile at the initial point A is $\left(2\hat{i} + 3\hat{j}\right)$ m/s. Its velocity (in m/s) at point B is

The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is