Select Chapter Topics:
Q. No.A projectile is launched with an initial speed of \(25~\text{m/s}\) at an angle \(\theta\) above the horizontal. After \(t\) seconds, its velocity becomes horizontal. If the horizontal range of the projectile is \(R,\) then what is the expression for the launch angle \(\theta\) in terms of \(R\) and \(t\)? (take \(g=10~\text{m/s}^2\))
1.
\(\dfrac{1}{2} \sin ^{-1}\left(\dfrac{5 t^2}{4 R}\right) \)
2.
\(\dfrac{1}{2} \sin ^{-1}\left(\dfrac{4 R}{5 t^2}\right) \)
3.
\(\tan ^{-1}\left(\dfrac{4 \mathrm{t}^2}{5 \mathrm{R}}\right) \)
4.
\(\cot ^{-1}\left(\dfrac{\mathrm{R}}{20 \mathrm{t}^2}\right)\)
Subtopic: Projectile Motion |
Level 3: 35%-60%
Please attempt this question first.
Hints
Please attempt this question first.
Given below are two statements:
| Assertion (A): | Two identical balls A and B thrown with the same velocity ‘u’ at two different angles with horizontal attained the same range R. If A and B reached the maximum height h1 and h2 respectively, then \(\mathrm{R}=4 \sqrt{\mathrm{h}_1 \mathrm{~h}_2}\) |
| Reason (R): | Product of said heights, \(\mathrm{h}_1 \mathrm{~h}_2=\left(\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}\right) \cdot\left(\frac{\mathrm{u}^2 \cos ^2 \theta}{2 \mathrm{~g}}\right)\) |
| 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. | (A) is false but (R) is true. |
Subtopic: Projectile Motion |
76%
Level 2: 60%+
JEE
Please attempt this question first.
Hints
Please attempt this question first.
For a particle in a uniform circular motion, the acceleration \(\vec a\) at any point P(R, \(\theta\)) on the circular path of radius R is: (when \(\theta\) is measured from the positive x-axis and v is uniform speed):
1. \(-\frac{v^2}{R} \sin \theta \hat{i}+\frac{v^2}{R} \cos \theta \hat{j} \)
2. \(-\frac{v^2}{R} \cos \theta \hat{i}+\frac{v^2}{R} \sin \theta \hat{j} \)
3. \(-\frac{v^2}{R} \cos \theta \hat{i}-\frac{v^2}{R} \sin \theta \hat{j} \)
4. \(-\frac{v^2}{R} \hat{i}+\frac{v^2}{R} \hat{j}\)
1. \(-\frac{v^2}{R} \sin \theta \hat{i}+\frac{v^2}{R} \cos \theta \hat{j} \)
2. \(-\frac{v^2}{R} \cos \theta \hat{i}+\frac{v^2}{R} \sin \theta \hat{j} \)
3. \(-\frac{v^2}{R} \cos \theta \hat{i}-\frac{v^2}{R} \sin \theta \hat{j} \)
4. \(-\frac{v^2}{R} \hat{i}+\frac{v^2}{R} \hat{j}\)
Subtopic: Circular Motion |
58%
Level 3: 35%-60%
JEE
Please attempt this question first.
Hints
Please attempt this question first.
A projectile is launched at an angle \('\alpha'\) with the horizontal with a velocity of 20 ms-1. After 10 s, its inclination with horizontal is \(' \beta'\). The value of \(tan \beta\) will be: (g = 10 ms-2)
1. \(\tan \alpha+5 \sec \alpha \)
2. \( \tan \alpha-5 \sec \alpha \)
3. \( 2 \tan \alpha-5 \sec \alpha \)
4. \(2 \tan \alpha+5 \sec \alpha\)
1. \(\tan \alpha+5 \sec \alpha \)
2. \( \tan \alpha-5 \sec \alpha \)
3. \( 2 \tan \alpha-5 \sec \alpha \)
4. \(2 \tan \alpha+5 \sec \alpha\)
Subtopic: Projectile Motion |
75%
Level 2: 60%+
JEE
Please attempt this question first.
Hints
Please attempt this question first.
Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly
A girl standing on the road holds her umbrella at \(45^{\circ}\) with the vertical to keep the rain away. If she starts running without an umbrella with a speed of \(15 \sqrt{2}~\text{km/h}, \) the raindrops hit her head vertically. The speed of raindrops with respect to the moving girl is:
1. \(30~\text{km/h} \)
2. \(\dfrac{25}{\sqrt{2}}~\text{km/h}\)
3. \(\dfrac{30}{\sqrt{2}}~\text{km/h}\)
4. \(25~\text{km/h} \)
1. \(30~\text{km/h} \)
2. \(\dfrac{25}{\sqrt{2}}~\text{km/h}\)
3. \(\dfrac{30}{\sqrt{2}}~\text{km/h}\)
4. \(25~\text{km/h} \)
Subtopic: Relative Motion |
Level 3: 35%-60%
Please attempt this question first.
Hints
Please attempt this question first.
Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly
The motion of a particle in the \(xy \text-\)plane is described by the following equations:
\(x=4 \sin \left(\dfrac{\pi}{2}-\omega t\right)~ \text m,\) \(y=4 \sin \left(\omega t\right) ~\text m.\)
Which of the following best describes the path traced by the particle?
\(x=4 \sin \left(\dfrac{\pi}{2}-\omega t\right)~ \text m,\) \(y=4 \sin \left(\omega t\right) ~\text m.\)
Which of the following best describes the path traced by the particle?
| 1. | circular | 2. | helical |
| 3. | parabolic | 4. | elliptical |
Subtopic: Circular Motion |
63%
Level 2: 60%+
Please attempt this question first.
Hints
Please attempt this question first.
An object is projected in the air with initial velocity \(u\) at an angle \(\theta\). The projectile motion is such that the horizontal range \(R\), is maximum. Another object is projected in the air with a horizontal range half of the range of the first object. The initial velocity remains the same in both cases. The value of the angle of projection, at which the second object is projected, will be:
| 1. | \(45^{\circ}\) | 2. | \(30^{\circ}\) |
| 3. | \(60^{\circ}\) | 4. | \(15^{\circ}\) |
Subtopic: Projectile Motion |
70%
Level 2: 60%+
Please attempt this question first.
Hints
Please attempt this question first.
A ball is projected with kinetic energy \(E\) at an angle of \(60^\circ\) above the horizontal. What will be its kinetic energy at the highest point of its trajectory?
| 1. | \(0\) | 2. | \(\dfrac{E}{2}\) |
| 3. | \(\dfrac{E}{4}\) | 4. | \(E\) |
Subtopic: Projectile Motion |
79%
Level 2: 60%+
Please attempt this question first.
Hints
Please attempt this question first.
Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly
A person can throw a ball up to a maximum range of \(100\) m. How high above the ground can he throw the same ball?
1. \(25\) m
2. \(50\) m
3. \(100\) m
4. \(200\) m
1. \(25\) m
2. \(50\) m
3. \(100\) m
4. \(200\) m
Subtopic: Projectile Motion |
Level 3: 35%-60%
Please attempt this question first.
Hints
Please attempt this question first.
Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly
Two inclined planes are placed as shown in the figure. A block is projected from the point \(A\) of the inclined plane \(AB\) along its surface with a velocity just sufficient to carry it to the top point \(B\) at a height of \(10~\text m.\) After reaching the point \(B\) the block slides down on the inclined plane \(BC.\) The time it takes to reach the point \(C\) from the point \(A\) is \(t( \sqrt 2 + 1 )~\text{s}.\) The value of \(t\) is:
(take \(g=10~\text{m/s}^2\))
1. \(1\)
2. \(2\)
3. \(3\)
4. \(4\)
(take \(g=10~\text{m/s}^2\))
1. \(1\)
2. \(2\)
3. \(3\)
4. \(4\)
Subtopic: Uniformly Accelerated Motion |
63%
Level 2: 60%+
Hints
Select Chapter Topics:
© 2025 GoodEd Technologies Pvt. Ltd.