(3) 45 m/sA shot is fired from a point at a distance of 200 mfrom the

1.	(3) 45 m/sA shot is fired from a point at a distance of 200 mfrom the foot of a tower 100 m high so that it justpasses over it horizontally. The direction of shot withhorizontal is35.(1) 30(3) 60°(2) 45°(4) 70°
Answer» For a projectile in 2-dimensions, we have the path equation as: y = x tanФ - gx² sec²Ф /2u² u = initial speed, Ф = angle of projection x = horizontal displacement, y = vertical displacement We assume that the tower height 100 m is the maximum height achieved by the bullet. Then the distance 200m will be equal to half of the range of the bullet.We know : H = u² Sin²Ф/2g 100 = u² Sin²Ф /(210)=> u² sin²Ф = 2,000 --- (1) R = u² Sin2Ф/g 2200 = u² Sin2Ф/10=> u² Sin2Ф = 4,000 --- (2) (2)÷ (1) => TanФ = 1 => Ф = 45°

(3) 45 m/sA shot is fired from a point at a distance of 200 mfrom the foot of a tower 100 m high so that it justpasses over it horizontally. The direction of shot withhorizontal is35.(1) 30(3) 60°(2) 45°(4) 70°

Answer»

For a projectile in 2-dimensions, we have the path equation as:

y = x tanФ - gx² sec²Ф /2u² u = initial speed, Ф = angle of projection x = horizontal displacement, y = vertical displacement

We assume that the tower height 100 m is the maximum height achieved by the bullet. Then the distance 200m will be equal to half of the range of the bullet.We know : H = u² Sin²Ф/2g 100 = u² Sin²Ф /(2*10)=> u² sin²Ф = 2,000 --- (1)

R = u² Sin2Ф/g 2*200 = u² Sin2Ф/10=> u² Sin2Ф = 4,000 --- (2)

(2)÷ (1) => TanФ = 1 => Ф = 45°

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