Q. No.The equation of the trajectory of a projectile is given as \(y = 4x-\dfrac{1}{4x^2}.\) The range of the projectile is:
1. 1 m
2. 32 m
3. 16 m
4. 48 m
1. \(1:1\)
2. \(2:1\)
3. \(1:2\)
4. \(1:3\)
A projectile is fired from the surface of the earth with a velocity of 5ms–1 and at an angle θ with the horizontal. Another projectile fired from another planet with a velocity of 3ms–1 at the same angle follows a trajectory that is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is: (given = 9.8 ms–2)
1. 3.5
2. 5.9
3. 16.3
4. 110.8
1. \(\sqrt 2 : 3\)
2. \(\sqrt 3: 2\)
3. \(4:9\)
4. \(2:1\)
| 1. | Their time of flight will be the same. |
| 2. | Their maximum height will be the same. |
| 3. | Their range will be the same. |
| 4. | Their landing velocity will be the same. |
The horizontal range is four times the maximum height attained by a projectile. The angle of projection is
1. 90
2. 60
3. 45
4. 30
Two bullets are fired horizontally with different velocities from the same height. Which will reach the ground first?
| 1. | slower one |
| 2. | faster one |
| 3. | both will reach simultaneously |
| 4. | cannot be predicted |
| 1. | \(25~\text{m/s} \) | 2. | \(40~\text{m/s}\) |
| 3. | \(31.25~\text{m/s}\) | 4. | \(12.5~\text{m/s}\) |
| Assertion (A): | When a body is projected at an angle of \(45^\circ,\) its range is maximum. |
| Reason (R): | For maximum range, the value of \(\text{sin}2 \theta\) should be equal to one. |
| 1. | Both (A) and (R) are true and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are true but (R) is not the correct explanation of (A). |
| 3. | (A) is true but (R) is false. |
| 4. | Both (A) and (R) are false. |
Ball \(A\) reaches the ground in:
1. \(8~\text s\)
2. \(4~\text s\)
3. \(2~\text s\)
4. \(\sqrt8~\text s\)
© 2026 GoodEd Technologies Pvt. Ltd.