An aeroplane is flying horizontally with a ve | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The horizontal velocity of the bomb is $u = 600 \, km/h = 600 	imes \frac{5}{18} \, m/s = \frac{500}{3} \, m/s$.The height from which the bomb is

Vedclass · Accepted Answer

$3.33 \, km$

Vedclass · Answer

(C) The horizontal velocity of the bomb is $u = 600 \, km/h = 600 	imes \frac{5}{18} \, m/s = \frac{500}{3} \, m/s$.The height from which the bomb is released is $h = 1960 \, m$.The time taken by the bomb to reach the ground is given by $t = \sqrt{\frac{2h}{g}}$.Taking $g = 9.8 \, m/s^2$,we have $t = \sqrt{\frac{2 	imes 1960}{9.8}} = \sqrt{\frac{3920}{9.8}} = \sqrt{400} = 20 \, s$.The horizontal distance $AB$ is the horizontal displacement,which is given by $AB = u 	imes t$.$AB = \frac{500}{3} 	imes 20 = \frac{10000}{3} \, m = 3333.33 \, m = 3.33 \, km$.

An aeroplane is flying horizontally with a velocity of $600 \, km/h$ at a height of $1960 \, m$. When it is vertically above a point $A$ on the ground,a bomb is released from it. The bomb strikes the ground at point $B$. The distance $AB$ is

Similar Questions

$A$ particle is projected with a velocity of $30\,m/s$ at an angle of $\theta_0 = \tan^{-1}(3/4)$. After $1\,s$,the particle is moving at an angle $\theta$ to the horizontal. What is the value of $\tan\theta$? (Take $g = 10\,m/s^2$)

$A$ ball is projected from the ground with a speed $15 \, m/s$ at an angle $\theta$ with the horizontal such that its range and maximum height are equal. Then,$\tan \theta$ will be equal to:

If a body projected with a velocity of $19.6 \ m/s$ reaches a maximum height of $9.8 \ m$,then the range of the projectile is (Neglect air resistance). (in $m$)

The horizontal range $(R)$ of a projectile is $n$ times its maximum height $(H)$. Find the angle of projection $(\theta_0)$.

$A$ bullet is fired at time $t=0$ with velocity $20 \ m/s$ and at an initial angle of $30^{\circ}$ with the horizontal. The angle between the displacement vector and the horizontal after time $0.1 \ s$ is (Assume $g=10 \ m/s^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module