Question

A bullet fired at an angle 30 degree with the horizonal hits the ground 3 km away . By adjusting it's angle of projection , can one hope to hits a target 5km away ? Assume the speed to be fix and naglect air resistance ( g= 10m/s)

Answer 1

Answer:

The bullet can not hit the target 5 km away.

Explanation:

we know,

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

Let the speed of the bullet be "u"

A bullet fired at an angle of 30° with the horizontal hits the ground 3 km away.

Range (R) = 3 km

angle of projection (θ) = 30°

• By substituting it's values in the above formula,

$R = \frac{ {u}^{2}sin2(30) }{g} \\ \\ 3= \frac{ {u}^{2} sin60}{g} \\ \\ 3= \frac{ {u}^{2} \times \frac{ \sqrt{3} }{2} }{g} \\ \\ 3= \frac{ \sqrt{3} {u}^{2} }{2g} \\ \\ \frac{ {u}^{2} }{g} = \frac{6}{ \sqrt{3} } \\ \\ \frac{ {u}^{2} }{g} = 2 \sqrt{3}$

• Maximum range is at θ = 45°

$R_{max} \: = \frac{ {u}^{2} sin2(45)}{g} \\ R_{max} = \frac{ {u}^{2}sin90 }{g} \\ R_{max} = \frac{ {u}^{2} }{g} \\ R_{max} = 2 \sqrt{3} \\ R_{max} = 2.8 \: km$

The bullet can not hit the target 5 km away.

Since the maximum range i.e., 2.8 km is less than 5km.