Question 1:
easy
Velocity versus time graph for a body projected vertically upwards is :-
Question 2:
easy
At what angle to the horizontal should an object be projected so that the maximum height reached is equal to the horizontal range?
Question 3:
easy
At what angle to the horizontal should an object be projected so that the maximum height reached is equal to the horizontal range?
Maximum height \(H = \frac{u^2\sin^2\theta}{2g}\) and range \(R = \frac{2u^2\sin\theta\cos\theta}{g}\). Equating the two yields \(\frac{\sin^2\theta}{2} = 2\sin\theta\cos\theta\), which simplifies to \(\tan\theta = 4\).
Question 4:
easy
At what angle to the horizontal should an object be projected so that the maximum height reached is equal to half of the horizontal range?
We require \(H = \frac{R}{2}\). Using the formulae \(H = \frac{u^2 \sin^2 \theta}{2g}\) and \(R = \frac{u^2 \sin 2\theta}{g}\), we get \(\frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \sin\theta\cos\theta}{g} ⇒ \tan\theta = 2 ⇒ \theta = \tan^{-1}(2)\).
Question 5:
easy
The range of a projectile, when launched at an angle of \(15^\circ\) with the horizontal is \(1.5\text{ km}\). What is the range of the projectile when launched at an angle of \(45^\circ\) to the horizontal?
Range is given by \(R = \frac{u^2 \sin(2\theta)}{g}\). For \(\theta = 15^\circ\), \(R_1 = \frac{u^2 \sin(30^\circ)}{g} = 1.5\text{ km}\) which implies \(\frac{u^2}{g} = 3.0\text{ km}\). For \(\theta = 45^\circ\), \(R_2 = \frac{u^2 \sin(90^\circ)}{g} = \frac{u^2}{g} = 3\text{ km}\).
Question 6:
easy
In projectile motion if air resistance (or any of such force opposing motion) is taken into consideration, then
Air resistance acts opposite to the direction of motion, decreasing velocity. This causes deviation from the ideal symmetric parabolic trajectory and decreases both the range and maximum height. Thus, (1) and (2) are correct.
Question 7:
easy
Assertion (A): Path of a projected ball is parabolic in uniform gravitational field for oblique projection in absence of air resistance.
Reason (R): Gravitational force is always act perpendicular to velocity during the motion of a projectile.
The path of a projectile is a parabola under constant gravitational acceleration. Gravitational force acts vertically downwards, which is perpendicular to the velocity only at the highest point of its trajectory, so R is false.
Question 8:
easy
Assertion (A): Path of a projected ball is parabolic in uniform gravitational field for oblique projection in absence of air resistance.
Reason (R): Gravitational force is always act perpendicular to velocity during the motion of a projectile.
Assertion (A) is true because projectile motion under gravity without air resistance follows a parabolic path.
Reason (R) is false as gravitational force acts perpendicular to velocity only at the peak of the trajectory, not always. Thus, (A) is true, (R) is false.
Question 9:
easy
Assertion (A): Horizontal component of velocity is constant in projectile motion under gravity.
Reason (R): Two projectiles having same horizontal range must have the same time of flight.
In projectile motion (neglecting air resistance), gravity acts only vertically. Thus, there is no horizontal acceleration, and the horizontal component of velocity remains constant. So (A) is true. Horizontal range is \(R = u_x T\). Projectiles launched at complementary angles have the same range but different times of flight (\(T = \frac{2u \sin\theta}{g}\)). So (R) is false.
Question 10:
easy
Assertion (A): Trajectory of an object moving under a constant acceleration is a straight line.
Reason (R): The shape of trajectory depends only on the acceleration.
An object under constant acceleration does not always follow a straight line (e.g., projectile motion is parabolic). A straight line occurs only if initial velocity is parallel or anti-parallel to acceleration. So (A) is false. The trajectory shape depends on both initial velocity and acceleration. So (R) is false.