The range of projectile is 1 .5 km when it is projected at an angle of 15° with horizontal. What will be its range when it is projected at an angle of 45° with the horizontal?
(A) 0.75 km
(B) 1.5 km
(C) 3 km
(D) 6 km
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Define Horizontal range of projectile.
The maximum horizontal distance travelled by the projectile is called the horizontal range (R) of the projectile.
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The range of projectile is 1 .5 km when it is projected at an angle of 15° with horizontal. What will be its range when it is projected at an angle
Correct Option is (C) 3 km We knoe that , range of projectile \(R = \frac{u^2 sin \ 2 \ \theta}{g}\) \(1.5 = \frac{u^2 sin(2 \times 15)}{g}\) \(1.5 = \frac{u^2}{2g}\) \(\frac{u^2}{g} = 3\) when, \(\theta = 45^\circ\) \(R...
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A player kicks up a ball at an angle θ with the horizontal. The horizontal range is maximum when θ is equal to
Correct Option is (B) \(45^\circ\) Range of projectile , \(R = \frac{u^2 \ sin2 \ \theta}{g}\) \(R_{max} = \frac{u^2 \ sin2 \ \theta}{g}\) \(\because \theta = 45^\circ\) \(R_{max} = \frac{u^2 \ sin 90^\circ}{g}\) \(R_{max} = \frac{u^2}{g}\)
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When air resistance is taken into account while dealing with the motion of the projectile, to achieve maximum horizontal range
Correct option is: (D) greater than 45°
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The maximum height attained by projectile is found to be equal to 0.433 of the horizontal range. The angle of projection of this projectile is
Correct Option is (C) \(60^\circ\) Given, H = 0.433R We know that , \(H = \frac{u^2sin^2 \theta}{2g} , R = \frac{u^2sin^2 \theta}{g}\) \( \frac{u^2sin^2 \theta}{g} = 0.433 \frac{u^2sin^2 \theta}{g}\) \(\frac{sin^2 \theta}{2g} = 0.433 . 2 \ sin...
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The maximum height attained by projectile is found to be equal to 0.433 of the horizontal range. The angle of projection of this projectile is
2 Answers
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A projectile is thrown with an initial velocity of 50 m/s. The maximum horizontal distance which this projectile can travel is
Correct Option is (B) 125m \(H_{max} = \frac{u^2}{2g}\) \(H_{max} = \frac{50 \times 50}{2 \times 10}\) \(H_{max} = 125 m\)
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A projectile is thrown with an initial velocity of 50 m/s. The maximum horizontal distance which this projectile can travel is
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A projectile projected with certain angle reaches ground with (A) double angle (B) same angle
Correct option is: (B) same angle
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A body is projected from ground with velocity u making an angle `theta` with horizontal. Horizontal range of body is four time to maximum height attai
Correct Answer - (i)`theta=45^(@)` (ii)`R=(u^(2))/(g),H=(u^(2))/(4g)` (iii) 1 : 1
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