A bullet fired at an angle of $30^o$ 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.
A bullet fired at an angle of $30^o$ 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.
No
Range, $R=3 \,km$ Angle of projection, $\theta=30^{\circ}$ Acceleration due to gravity, $g=9.8 \,m / s ^{2}$
Horizontal range for the projection velocity $u_{0}$, is given by the relation:
$R=\frac{u_{0}^{2} \sin 2 \theta}{g}$
$3=\frac{u_{0}^{2}}{g} \sin 60^{\circ}$
$\frac{u_{0}^{2}}{g}=2 \sqrt{3}$
The maximum range $\left(R_{\max }\right)$ is achieved by the bullet when it is fired at an angle of $45^{\circ}$ with the horizontal, that is, $R_{\max }=\frac{u_{0}^{2}}{g}$
On comparing above equations , we get:
$R_{\max }=3 \sqrt{3}=2 \times 1.732=3.46 \,km$
Hence, the bullet will not hit a target $5 \,km$ away.
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$(b)$ Show that both $\widehat r$ and $\widehat \theta $ are unit vectors and are perpendicular to each other.
$(c)$ Show that $\frac{d}{{dr}}(\widehat r)\, = \,\omega \hat \theta \,$, where $\omega \, = \,\frac{{d\theta }}{{dt}}$ and $\frac{d}{{dt}}(\widehat \theta )\, = \, - \theta \widehat r\,$.
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