A bullet fired at an angle of 30° with the horizontal hits the ground

1.	A bullet fired at an angle of 30° with the horizontal hits the ground 4.0 km away. Byadjusting its angle of projection, can one hope to hit a target 5.0 km away ? Assumethe muzzle speed to be fixed, and neglect air resistance.
Answer» Answer : Bullet can not hit a TARGET 5 km away. Step-by-step explanation : we know, $\boxed{\bf Range=\frac{u^2sin2\theta}{g} }$ LET the speed of the bullet be "u" A bullet fired at an angle of 30° with the HORIZONTAL hits the ground 4.0 km away. Range (R) = 4 km angle of projection (θ) = 30° By SUBSTITUTING it's values in the above formula, $4=\frac{u^2sin2(30^{\circ})}{g} \\\\ 4\frac{u^2sin60^{\circ}}{g} \\\\ \frac{u^2(\frac{\sqrt{3}}{2})}{g} =4 \\\\ \frac{u^2}{g} =\frac{8}{\sqrt{3}} \\\\ \frac{u^2}{g} =\frac{8}{\sqrt{3}} \times \frac{\sqrt{3} }{\sqrt{3} } \\\\ \frac{u^2}{g} =\frac{8\sqrt{3}}{3}$ Range is maximum at θ = 45° $R_{max}=\frac{u^2sin2(45^{\circ})}{g} \\\\ R_{max} =\frac{u^2sin90^{\circ}}{g} \\\\ R_{max}=\frac{u^2}{g}$ substitute the value of u²/g here, Rₘₐₓ = $\frac{8\sqrt{3}}{3}$ $\simeq\frac{13.856}{3} \\\\ \simeq\text{4.618 km}$ Hence, bullet can not hit the target 5 km away. SINCE, the maximum range with the same speed is 4.618 km which is less than 5 km _______________________

A bullet fired at an angle of 30° with the horizontal hits the ground 4.0 km away. Byadjusting its angle of projection, can one hope to hit a target 5.0 km away ? Assumethe muzzle speed to be fixed, and neglect air resistance.

Answer»

Answer :

Bullet can not hit a TARGET 5 km away.

Step-by-step explanation :

we know,

$\boxed{\bf Range=\frac{u^2sin2\theta}{g} }$

LET the speed of the bullet be "u"

A bullet fired at an angle of 30° with the HORIZONTAL hits the ground 4.0 km away.

Range (R) = 4 km

angle of projection (θ) = 30°

By SUBSTITUTING it's values in the above formula,

$4=\frac{u^2sin2(30^{\circ})}{g} \\\\ 4\frac{u^2sin60^{\circ}}{g} \\\\ \frac{u^2(\frac{\sqrt{3}}{2})}{g} =4 \\\\ \frac{u^2}{g} =\frac{8}{\sqrt{3}} \\\\ \frac{u^2}{g} =\frac{8}{\sqrt{3}} \times \frac{\sqrt{3} }{\sqrt{3} } \\\\ \frac{u^2}{g} =\frac{8\sqrt{3}}{3}$

Range is maximum at θ = 45°

$R_{max}=\frac{u^2sin2(45^{\circ})}{g} \\\\ R_{max} =\frac{u^2sin90^{\circ}}{g} \\\\ R_{max}=\frac{u^2}{g}$

substitute the value of u²/g here,

Rₘₐₓ = $\frac{8\sqrt{3}}{3}$

$\simeq\frac{13.856}{3} \\\\ \simeq\text{4.618 km}$

Hence, bullet can not hit the target 5 km away.

SINCE, the maximum range with the same speed is 4.618 km which is less than 5 km

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