A bullet fired at an angle of 30^{circ} with | vedclass

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

Question

(B) The horizontal range $R$ of a projectile is given by the formula: $R = \frac{u^2 \sin(2	heta)}{g}$,where $u$ is the muzzle speed,$	heta$ is the

Vedclass · Accepted Answer

No

Vedclass · Answer

(B) The horizontal range $R$ of a projectile is given by the formula: $R = \frac{u^2 \sin(2	heta)}{g}$,where $u$ is the muzzle speed,$	heta$ is the angle of projection,and $g$ is the acceleration due to gravity.Given: $R_1 = 3.0 \; km$ at $	heta_1 = 30^{\circ}$.Substituting these values: $3.0 = \frac{u^2 \sin(60^{\circ})}{g} = \frac{u^2}{g} \cdot \frac{\sqrt{3}}{2}$.Thus,$\frac{u^2}{g} = \frac{3.0 	imes 2}{\sqrt{3}} = 2\sqrt{3} \approx 3.464 \; km$.The maximum range $R_{	ext{max}}$ for a fixed muzzle speed $u$ occurs at $	heta = 45^{\circ}$,where $R_{	ext{max}} = \frac{u^2}{g}$.Substituting the value of $\frac{u^2}{g}$: $R_{	ext{max}} = 2\sqrt{3} \approx 3.464 \; km$.Since the maximum possible range is approximately $3.46 \; km$,it is impossible to hit a target at $5.0 \; km$ by only adjusting the angle of projection.

Column $I$	Column $II$
$(A)$ Displacement	$(p)$ $8 \sin 2$
$(B)$ Distance	$(q)$ $4$
$(C)$ Average velocity	$(r)$ $2 \sin 2$
$(D)$ Average acceleration	$(s)$ $4 \sin 2$

$A$ bullet fired at an angle of $30^{\circ}$ with the horizontal hits the ground $3.0 \; km$ away. By adjusting its angle of projection,can one hope to hit a target $5.0 \; km$ away? Assume the muzzle speed to be fixed,and neglect air resistance.

Similar Questions

The coordinates of a particle moving in the $x-y$ plane are given by: $x = 2 + 4t$,$y = 3t + 8t^2$. The motion of the particle is:

$A$ particle of mass $m$ is performing $U.C.M.$ along a circle of radius $r$. The relation between centripetal acceleration $a$ and kinetic energy $E$ is given by

$A$ ball is projected from the ground at an angle of $\theta$ from the horizontal. The graph of kinetic energy $(K)$ versus time $(t)$ will be:

The position vector of a particle is given as $\vec{r} = (t^2 - 4t + 6)\hat{i} + (t^2)\hat{j}$. The time after which the velocity vector and acceleration vector become perpendicular to each other is equal to ....... $sec$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module