A bomb is dropped by an airplane flying horiz | vedclass

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

Question

(C) The airplane is flying horizontally,so the initial vertical component of the bomb's velocity is $u_y = 0 	ext{ m/s}$.Using the equation of motion

Vedclass · Accepted Answer

$\frac{10^4}{9 \sqrt{2}} 	ext{ m}$

Vedclass · Answer

(C) The airplane is flying horizontally,so the initial vertical component of the bomb's velocity is $u_y = 0 	ext{ m/s}$.Using the equation of motion for the vertical direction: $h = \frac{1}{2} gt^2$.Substituting the given values: $980 = \frac{1}{2} 	imes 9.8 	imes t^2$.$t^2 = \frac{980 	imes 2}{9.8} = 100 	imes 2 = 200$.$t = \sqrt{200} = 10\sqrt{2} 	ext{ s}$.The horizontal velocity of the bomb is $v_x = 200 	ext{ km/hr} = 200 	imes \frac{5}{18} = \frac{1000}{18} 	ext{ m/s}$.The horizontal distance $d$ covered by the bomb during its fall is $d = v_x 	imes t$.$d = \frac{1000}{18} 	imes 10\sqrt{2} = \frac{10000\sqrt{2}}{18} = \frac{5000\sqrt{2}}{9} 	ext{ m}$.Wait,simplifying the expression: $d = \frac{1000}{18} 	imes 10\sqrt{2} = \frac{10000}{9\sqrt{2}} 	ext{ m}$.

Column-$I$	Column-$II$
$(1)$ Vertical component of velocity is zero	$(a)$ Tangent to the parabolic path
$(2)$ Linear velocity	$(b)$ Maximum height point of the projectile trajectory

$A$ bomb is dropped by an airplane flying horizontally with a velocity of $200 \text{ km/hr}$ at a height of $980 \text{ m}$. At the time of dropping the bomb,the horizontal distance of the airplane from the target on the ground to hit it directly is (given $g = 9.8 \text{ m/s}^2$):

Similar Questions

$A$ particle is projected from the ground with a kinetic energy $E$ at an angle of $60^{\circ}$ with the horizontal. Its kinetic energy at the highest point of its motion will be

Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ Vertical component of velocity is zero $(a)$ Tangent to the parabolic path

$(2)$ Linear velocity $(b)$ Maximum height point of the projectile trajectory

$A$ cricket player hits a ball like a projectile,and the fielder catches the ball after $2 \ s$. The maximum height reached by the ball is $(g = 10 \ m/s^2)$. (in $m$)

For a projectile motion, the range $R$ is '$n$' times the maximum height $H$. So the angle of projection is

$A$ projectile is projected from the ground with an initial velocity $\vec{u} = u_0 \hat{i} + v_0 \hat{j}$. If the acceleration due to gravity $g$ is along the negative $y$-direction,find the maximum displacement in the $x$-direction (horizontal range).

Vedclass Test Series

Exam Paper Generator

Online Exam Module