Assertion (A): A particle is projected from ground on a horizontal plane with speed \(10\text{ ms}^{-1}\) and angle of projection \(37^\circ\) with horizontal. Its velocity vector will be perpendicular to initial velocity vector after \(\frac{4}{3}\text{ s}\).
Reason (R): Two vectors \(\vec{v}\) and \(\vec{u}\) are perpendicular then \(\vec{u} \cdot \vec{v} = 0\).
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A): Initial velocity \(\vec{u} = (10 cos 37^\circ)\hat{i} + (10 sin 37^\circ)\hat{j} approx 8\hat{i} + 6\hat{j}\). Velocity at time (t) is \(\vec{v}(t) = 8\hat{i} + (6 - gt)\hat{j}\). For perpendicularity, \(\vec{u} \cdot \vec{v} = 0 ⇒ 64 + 6(6 - gt) = 0 ⇒ 100 - 6gt = 0\). With \(g=10\text{ m/s}^2), (t = 100/60 = 5/3\text{ s}\). The assertion states \(4/3\text{ s})\, so (A) is False.
Reason (R): The dot product of two perpendicular vectors is indeed zero. So (R) is True.
Since (A) is false and (R) is true, none of the given options are strictly correct. However, if (A) is false, options (1), (2), (3) are ruled out, leaving (4) by elimination, despite (R) being true.
Assertion (A): In projectile motion (from ground to ground projection), horizontal range is always same for angle of projection \(\theta\) and \(90^\circ – \theta\).
Reason (R): Horizontal range is independent of angle of projection.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A): The range formula is \(R = \frac{u^2 sin(2\theta)}{g}\). For angle \((90^\circ - \theta)\), \(R' = \frac{u^2 sin(2(90^\circ - \theta))}{g} = \frac{u^2 sin(180^\circ - 2\theta)}{g} = \frac{u^2 sin(2\theta)}{g} = R\). So (A) is True.
Reason (R): The horizontal range clearly depends on the angle of projection (theta) via (sin(2theta)). So (R) is False.
Since (A) is true and (R) is false, option (3) is correct.
Assertion (A): In projectile motion, speed always decreases.
Reason (R): In presence of air drag, projectile motion is a uniformly accelerated motion.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A): In ideal projectile motion, speed decreases on the way up and increases on the way down, reaching a minimum at the peak. It does not always decrease. So (A) is False.
Reason (R): In the presence of air drag, the drag force depends on velocity, making the net acceleration non-constant. Thus, it is not uniformly accelerated motion. So (R) is False.
Since both (A) and (R) are false, option (4) is correct.
A particle is projected with a speed of \( 20 \, \text{m s}^{-1} \) from level ground at an angle \( \theta \) equal to \( 45^\circ \) from horizontal. The ratio of maximum height attained by the body to horizontal range acquired by the body will be
1. \( \frac{2}{1} \)
2. \( \frac{1}{2} \)
3. \( \frac{1}{4} \)
4. \( \frac{3}{2} \)
View Answer
The ratio of maximum height \( H \) to horizontal range \( R \) is given by \( \frac{H}{R} = \frac{\tan \theta}{4} \). For \( \theta = 45^\circ \), \( \tan 45^\circ = 1 \), leading to \( \frac{H}{R} = \frac{1}{4} \).