Projectile



  • A building 4.8 m high 2b meters wide has a flat roof. A ball is projected from a point on the horizontal
    ground 14.4 m away from the building along its width. If projected with velocity 16 m/s at an angle of 45º
    with the ground, the ball hits the roof in the middle, find the width 2b. Also find the angle of projection so that
    the ball just crosses the roof if projected with velocity 10 3 m/s.(g=10m/s2)



  • @Manas-Sontakke

    Let R be the horizontal distance covered by the projectile then and a be the angle of projection.
    U be the initial velocity and in our case it is 16m/s
    And g be the acceleration due to gravity and in our case it is 10.
    ​​
    Then R = U² Sin 2a / g
    R = 14.4 + b (it hits the roof half way and the width of the roof is 2b)
    R = 16² sin 2(45°) / 10
    = (256 × 0.98769) / 10 = 25.28m
    b = 25.28 - 14.4 = 10.88m
    2b = 2 × 10.88 = 21.76m
    This is the width of the building.

    The angle of projection when initial velocity is 10.3m/s and the projectile crosses the roof is given by :
    Height of the roof equals U Sin a where u is the initial velocity.
    4.8 = 10.3 × Sin a
    Sin a = 4.8 ÷ 10.3 = 0.4660
    Sin ⁻¹0. 4660 = 30.86°


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