Equation Of Path Of A Projectile MCQ - Practice Questions with Answers

A particle just clears a wall of height b at a distance a and strikes the ground at a distance c from the point of projection. The angle of projection is:

An object is projected with a velocity of  making an angle of   with horizontal The equation for the trajectory is  is height    is horizontal distance, and  are constant, the ratio  is

A body of mass  is being Projected with an angle of  with the vertical. The trajectory of body is observed at point  (30,20) If the '  ', is time of fight Then its momentum vector, at time  is

If the initial velocity in the Horizontal direction of a projectile is unit vector $\hat{i}$ and the equation of trajectory is $\mathrm{y}=10 \mathrm{x}(1-\mathrm{x})$. Find the y -component vector of initial velocity $\left(g=10 \mathrm{~m} / \mathrm{sec}^2\right)$

Assertion: For $x=10 m ;  y=12 x-\frac{6}{10} x^2$, this projectile motion the equation will reach the maximum height

Reason: Maximum height depends upon the initial velocity of the projectile.

For a parabolic path of projection $y=x-\frac{g x^2}{2 u^2 \cos ^2 \Theta}$. The angle of projection is 

The equation of projectile is given as $y=x-\frac{g x^2}{10}$. Then the maximum range of the projectile is ( x and y are in meters)