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A bullet is fired from a gun at the speed of \(280~\text{ms}^{-1}\) in the direction \(30^\circ\) above the horizontal. The maximum height attained by the bullet is: \(\left(g=9.8~\text{m/s}^{2}, \sin30^{\circ}=0.5\right)\)
1. \(3000~\text{m}\) 2. \(2800~\text{m}\)
3. \(2000~\text{m}\) 4. \(1000~\text{m}\)
Q1:
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Subtopic:  Projectile Motion |
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A ball is projected from point \(A\) with velocity \(20\) ms–1 at an angle \(60^\circ\) to the horizontal direction. At the highest point \(B\) of the path (as shown in figure), the velocity \(v\) (in ms–1) of the ball will be:
  
1. \(20\)
2. \(10\sqrt3\)
3. zero
4. \(10\)
Q2:
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Subtopic:  Projectile Motion |
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A ball is projected with a velocity of \(10\) ms–1 at an angle of \(60^\circ\) with the vertical direction. Its speed at the highest point of its trajectory will be:
1. \(10\) ms–1
2. zero
3. \(5\sqrt3\) ms–1
4. \(5\) ms–1
Q3:
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Subtopic:  Projectile Motion |
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A cricket ball is thrown by a player at a speed of \(20\) m/s in a direction \(30^\circ\) above the horizontal. The maximum height attained by the ball during its motion is: (Take \(g=10\) m/s2)
1. \(5\) m
2. \(10\) m
3. \(20\) m
4. \(25\) m
Q4:
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Subtopic:  Projectile Motion |
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A car starts from rest and accelerates at \(5~\text{m/s}^{2}\). At \(t=4~\text{s}\), a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at \(t=6~\text{s}\)? (Take \(g=10~\text{m/s}^2)\)
1. \(20\sqrt{2}~\text{m/s}, 0~\text{m/s}^2\)
2. \(20\sqrt{2}~\text{m/s}, 10~\text{m/s}^2\)
3. \(20~\text{m/s}, 5~\text{m/s}^2\)
4. \(20~\text{m/s}, 0~\text{m/s}^2\)

Q5:
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Subtopic:  Projectile Motion |
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A particle moving in a circle of radius \(R\) with a uniform speed takes a time \(T\) to complete one revolution. If this particle were projected with the same speed at an angle \(\theta\) to the horizontal, the maximum height attained by it equals \(4R.\) The angle of projection, \(\theta\) is then given by:
1. \( \theta=\sin ^{-1}\left(\dfrac{\pi^2 {R}}{{gT}^2}\right)^{1/2}\) 2. \(\theta=\sin ^{-1}\left(\dfrac{2 {gT}^2}{\pi^2 {R}}\right)^{1 / 2}\)
3. \(\theta=\cos ^{-1}\left(\dfrac{{gT}^2}{\pi^2 {R}}\right)^{1 / 2}\) 4. \(\theta=\cos ^{-1}\left(\dfrac{\pi^2 {R}}{{gT}^2}\right)^{1 / 2}\)
Q6:
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Subtopic:  Projectile Motion |
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