Question

The maximum range of a gun of horizontal terrain is 16 km.if g=10ms-2, then muzzle velocity of a shell must be

Answer 1

Maximum range will be when the gun is shot at 45°.

Proof is as follows :

Assume a gun is shot at angle Ф with horizontal velocity V.

U(initial velocity) = (V Cos Ф) i + (V sin Ф) j

S = ut + 0.5at²

a = - 10j (since as the body is projected at an angle it experiences deceleration)

S = (V. Cos Фt) i + (V. Sin Фt - 5t²) j

Final y coordinate will be 0.

So :

t = V. Sin Ф / 5

Range = V. Cos Фt

= V. Cos Ф. Sin Ф / 5

= V²Sin 2Ф /10

Maximum range will be there when Sin 2Ф = 1 or Ф = 45°

Maximum range is thus :

V²/10 = 16 × 1000

V² / 10 = 16000

V² = 160000

V = √160000

V = 400m/s

Answer 2

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Answer:

400 m/s

Explanation:

The maximum range is at 45°

As we know,

Range(R) = u^2 Sin2⊙/g. -----------1

Given - Range = 16km =16000 m/s

g= 10m/s^2

Putting values in equation 1

16000 =u^2 Sin2(45°)/10

16000*10=u^2 Sin 90°

160000=u^2 *1 [Sin 90°=1]

160000=u^2

Taking square root on both sides

we get,

u=400m/s

Note: -- {⊙ =angle or theta}

Hope it helps..