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3 May, 14:41

g A cannonball is shot with an initial speed of 62 meters per second at a launch angle of 25 degrees toward a castle wall that is 260 meters away. If the wall is 20 meters tall, how high off the ground will the cannonball hit

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  1. 3 May, 15:10
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    h = 16.23 m

    The cannonball will hit the wall at 16.23m from the ground.

    Step-by-step explanation:

    Given;

    Initial speed v = 62m/s

    Angle ∅ = 25°

    Horizontal distance d = 260 m

    Height of wall y = 20

    Resolving the initial speed to vertical and horizontal components;

    Horizontal vx = vcos∅ = 62cos25°

    Vertical vy = vsin∅ = 62cos25°

    The time taken for the cannon ball to reach the wall is;

    Time t = horizontal distance/horizontal speed

    t = d/vx (since horizontal speed is constant)

    t = 260 / (62cos25°)

    t = 4.627 seconds.

    Applying the equation of motion;

    The height of the cannonball at time t is;

    h = (vy) t - 0.5gt^2

    Acceleration due to gravity g = 9.81 m/s

    Substituting the given values;

    h = 62sin25*4.627 - 0.5*9.81*4.627^2

    h = 16.2264134736

    h = 16.23 m

    The cannonball will hit the wall at 16.23m from the ground.
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