Question

A bullet is fired at an angle of 30° with the horizantal hits the ground 3√3 km away.Can we hit a target at a distance of 6√2 km by adjusting its angle of projection?

Answer 1

We know,

Range = u²sin2∅/g

Range is maximum at ∅ = 45°

e.g maximum range = u²/g

A/C to question ,

Range = u²sin2×30°/g

3000 m = u²×√3/2/g

20000√3 = u² -------(1)

Put in maximum range

Maximum range = (20000√3)/10

= 2000√3 = 2√3 km = 3.464 km

Hence, bullet can't be fired up to 5km { due to maximum range = 3.464 km } with the same speed .

Answer 2

Answer:

No.

Explanation:

Range of the bullet in first case is 3√3 km, ø = 30°

R = u^2sin2ø /g

3√3 = u^2(sin60°)/g

3√3 = u^2(√3)/2g

u^2 = 6g

Maximum range would be covered by the particle when ø = 45°

R = u^2sin90°/g

R = 6g(1)/g

R = 6km

Hence the maximum range of the particle even after altering the range is 6km. You can't hit a target at the distance of 6√2 km

Note:

• I took range in 'km', you can convert it to SI unit too.
• If you have any doubt reg why I took ø = 45° for maximum range, remember sinø is maximum at 90°, so in case of range, sin2ø, 2ø = 90°, ø = 45°.