A projectile projected with certain angle reaches ground with
(A) double angle
(B) same angle
(C) greater than 90°
(D) angle between 90° and 180°
1 Answers
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A proton X is projected in a region of uniform electric field E with speed u. Another proton Y is projected simultaneously with same speed u in a sepa
Correct Answer - (a) `(pi)/(2) + sin^(-1) ((sigmaET)/(2u))` , (b) `sqrt(u^(2) - (sigma^(2)E^(2)T^(2))/(4)) .T`
2 Answers 1 views
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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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
2 Answers
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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 \(H_{max} = \frac{u^2}{2g}\) \(H_{max} = \frac{50 \times 50}{2 \times 10}\) \(H_{max} = 125 m\)
2 Answers 8 views
A projectile is thrown with an initial velocity of 50 m/s. The maximum horizontal distance which this projectile can travel is
2 Answers
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A projectile projected with certain velocity reaches ground with (magnitude)
Correct option is: (C) same velocity
2 Answers 3 views
A body projected vertically upwards, reaches a height of 180 m. Find out `[g=10ms^(-2)]` (i) intial velocity of body (ii) time taken by body to reach
Correct Answer - (i)`60m//s` (ii) 6 sec (iii) 12 sec (iv )160 m (v)`-20m//s` (vi) 2 sec, 10 sec (vii) 6 sec
2 Answers 1 views
A ball is projected from ground in such a way that after 10 seconds of projection it lands on ground 500 m away from the point of projection. Find out
Correct Answer - (i)`theta=45^(@)` (ii)`50sqrt(2)m//s` (iii)`50m//sec`
2 Answers 10 views