Q1. In projectile motion, the horizontal range is maximum when the angle of projection is:
Correct Answer: (b) 45°
The horizontal range \( R = \frac{u^2 \sin 2\theta}{g} \) is maximum when \( \sin 2\theta = 1 \), which occurs at \( \theta = 45° \).
Q2. The time of flight of a projectile depends on:
Correct Answer: (b) Vertical velocity only
The time of flight \( T = \frac{2u \sin \theta}{g} \) depends only on the vertical component of velocity.
Q3. At the highest point of a projectile's trajectory, the:
Correct Answer: (c) Kinetic energy is minimum
At the highest point, the vertical velocity becomes zero but horizontal velocity remains, making kinetic energy minimum (but not zero).
Q4. The horizontal range of a projectile is equal to its maximum height when the angle of projection is:
Correct Answer: (d) 76°
When \( R = H \), we have \( \frac{u^2 \sin 2\theta}{g} = \frac{u^2 \sin^2 \theta}{2g} \). Solving gives \( \tan \theta = 4 \) or \( \theta \approx 76° \).
Q5. For a projectile, the ratio of maximum height to horizontal range when θ = 45° is:
Correct Answer: (c) 1:4
At \( \theta = 45° \), \( H = \frac{u^2 \sin^2 45°}{2g} = \frac{u^2}{4g} \) and \( R = \frac{u^2 \sin 90°}{g} = \frac{u^2}{g} \). Thus \( H/R = 1/4 \).
Q6. The path of a projectile is a:
Correct Answer: (c) Parabola
Under uniform gravity and no air resistance, the trajectory follows a parabolic path described by \( y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta} \).
Q7. If the initial speed of a projectile is doubled, its maximum height becomes:
Correct Answer: (b) 4 times
Since maximum height \( H \propto u^2 \), doubling the initial speed \( u \) makes \( H \) increase by a factor of 4.
Q8. Two projectiles are fired at angles 30° and 60° with the same speed. Their horizontal ranges are:
Correct Answer: (b) Same
The range \( R = \frac{u^2 \sin 2\theta}{g} \) is the same for complementary angles (θ and 90°-θ) when projected with the same speed.
Q9. The maximum height attained by a projectile is half its horizontal range. The angle of projection is:
Correct Answer: (a) 30°
When \( H = R/2 \), we get \( \frac{u^2 \sin^2 \theta}{2g} = \frac{1}{2} \times \frac{u^2 \sin 2\theta}{g} \). Simplifying gives \( \tan \theta = 1/2 \) or \( \theta \approx 26.6° \). (Note: The closest option is 30°)
Q10. A projectile is fired at an angle θ with speed u. The time taken to reach the highest point is:
Correct Answer: (a) \( \frac{u \sin \theta}{g} \)
The time to reach maximum height depends only on the vertical component: \( t = \frac{u \sin \theta}{g} \).