A projectile is thrown at an angle 15• with the horizontal and the range is 1.5km.what is the range when it is projected at 45•?

Question

Answer

R = u² * (sin 2∅)/ghere ∅=15°==> 1.5 = (u²/g) * sin 2*15°==> u²/g = 1.5 / sin 30°{as sin 30° = 1/2   ==>   1/ sin 30= 2}therefore==> u²/g = 1.5*2. = 3. ......{1}now R at 45 ° -->==> R = (u²/g) * ( sin 2*45°)==> R = (u²/g) * ( sin 90° ){as  ( sin 90° )= 1 } => R = ( u²/g ) = 3 km     from {1}

Answer

3

Answer

We know range = U×U ×2sin (angle)/g
so=range = sin3o/sin90 =1.5/R2 (new range Let R2)
so we get finally answer 3Km...

Answer

how?

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