A shot is fired from a point at a distance of 200m from the foot of a towre 100m hight so that it just passes over it. find the magnitude and diretion of velocity of the shot
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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°
=> u² = 4000
u = 20√10 m/s
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°
=> u² = 4000
u = 20√10 m/s
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