A stone projected with a velocity u at an angle θ with the horizontal reaches maximum height H1. When it is projected with velocity u at an angle π2θwith the horizontal, it reaches maximum height H2. The relation between the horizontal range of the projectile R and H1 & H2 is: 

1. \(R=4 \sqrt{H_1 H_2} \) 2. \(R=4\left(H_1-H_2\right) \)
3. \(R=4\left(H_1+H_2\right) \) 4. \(R=\frac{H_1{ }^2}{H_2{ }^2}\)

Subtopic:  Projectile Motion |
 83%
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Which of the following sets of factors will affect the horizontal distance covered by an athlete in a long–jump event?
1. speed before he jumps and his weight
2. the direction in which he leaps and the initial speed
3. the force with which he pushes the ground and his speed
4. none of the above

Subtopic:  Projectile Motion |
 62%
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In a projectile motion, velocity at maximum height is 

(1) ucosθ2

(2) ucosθ

(3) usinθ2

(4) None of these

Subtopic:  Projectile Motion |
 84%
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The equation of motion of a projectile is given by x = 36 t metre and 2y = 96 t – 9.8 t2 metre. The angle of projection is:

1. sin145

2. sin135

3. sin143

4. sin134

Subtopic:  Projectile Motion |
 63%
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For a given velocity, a projectile has the same range of R for two angles of projection. If t1 and t2 are the times of flight in the two cases then:

(1) t1t2R2

(2) t1t2R

(3) t1t21R

(4) t1t21R2

Subtopic:  Projectile Motion |
 67%
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A body of mass m is thrown upwards at an angle θ with the horizontal with velocity v. While rising up the velocity of the mass after t seconds will be 

(1) (vcosθ)2+(vsinθ)2

(2) (vcosθvsinθ)2gt

(3) v2+g2t2(2vsinθ)gt

(4) v2+g2t2(2vcosθ)gt 

Subtopic:  Projectile Motion |
 63%
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A cricketer can throw a ball to a maximum horizontal distance of 100 m. With the same effort, he throws the ball vertically upwards. The maximum height attained by the ball is 

1. 100 m

2. 80 m

3. 60 m

4. 50 m

Subtopic:  Projectile Motion |
 75%
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A ball is projected with velocity v0 at an angle of elevation 30°. Mark the correct statement.

(1) Kinetic energy will be zero at the highest point of the trajectory.

(2) Vertical component of momentum will be conserved.

(3) Horizontal component of momentum will be conserved.

(4) Gravitational potential energy will be minimum at the highest point of the trajectory.

Subtopic:  Projectile Motion |
 78%
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PMT - 2004
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Neglecting the air resistance, the time of flight of a projectile is determined by:

(1) Uvertical

(2) Uhorizontal

(3) U = U2vertical + U2horizontal

(4) U = U(U2vertical + U2horizontal )1/2

Subtopic:  Projectile Motion |
 74%
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A stone is thrown at an angle θ with the horizontal reaches a maximum height of H. Then the time of flight of stone will be:

(1) 2Hg

(2) 22Hg

(3) 22Hsinθg

(4) 2Hsinθg

Subtopic:  Projectile Motion |
 54%
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