Projectile Basics & Trajectory

The equation of the trajectory of a ball projected at an angle $\theta$ with the horizontal, is given as $y = x - \frac{gx^{2}}{2}$

The initial velocity of the ball is

[Given: $\tan 45^\circ = 1,\ \cos 45^\circ = \frac{1}{\sqrt{2}}$ ]

The trajectory of projectile, projected from the ground is given by $y = x - \frac{x^{2}}{20}$. Where $x$ and $y$ are measured in meter. The maximum height attained by the projectile will be.

If the initial velocity in horizontal direction of a projectile is unit vector $\hat{i}$ and the equation of trajectory$y = 5x(1 - x)$. The $y$ component vector of the initial velocity is ______________ $\hat{j}$. ($\text{Take } g = 10\text{ m/s}^{2})$

The trajectory of a projectile in a vertical plane is y = $\alpha$x $-\beta$x2, where $\alpha$ and $\beta$ are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection $\theta$ and the maximum height attained H are respectively given by:

The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x2. If it were launched at an angle $\theta$0 with speed v0 then (g = 10 ms–2):

The position of a projectile launched from the origin at t = 0 is given by$\vec{r}$= (40 $\hat{i}$ + 50 $\hat{j}$) m at t = 2s. If the projectile was launched at an angle $\theta$ from the horizontal, then $\theta$ is (take g = 10 ms${}^{-2}$).

A projectile is given an initial velocity of $\left(\hat{i} + 2\hat{j}\right)$ m/s, where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. If g = 10 m/s2, the equation of its trajectory is:

A ball projected from ground at an angle of 45° just clears a wall in front. If point of projection is 4 m from the foot of wall and ball strikes the ground at a distance of 6 m on the other side of the wall, the height of the wall is:

A projectile is fired at an angle of 45o with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection, is

A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an angle of $30^\circ$ with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground? $\left[g = 10\text{ m/s}^{2},\ \sin 30^\circ = \frac{1}{2},\ \cos 30^\circ = \frac{\sqrt{3}}{2}\right]$