1. A projectile is fired with an initial speed of 75.2m/s at an angle of 34.5degree

above the horizontal on a long flat firing range. What is the velocity of the

projectile 1.5s after firing?

1 Answer

  • 1 decade ago
    Favorite Answer

    Vertical motion and horizontal motion are independent.Firstly, consider the horizontal motion, the projectile travels at constant horizontal speed.

    Horizontal speed, vh = 75.2cos34.5* = 61.974 ms-1

    Lastly, we consider the vertical motion, take upward direction be positive.

    Initial speed, u = 75.2sin34.5 ms-1

    Time of flight, t = 1.5 s

    Acceleration due to gravity, g = -10 ms-2

    By v = u + gt

    v = 75.2sin34.5* + (-10)(1.5)

    vv = 27.594 ms-1

    Therefore, the resultant speed

    = √(vh2 + vv2)

    = √[61.9742 + 27.5942]

    = 67.8 ms-1

    Let x be the angle between the velocity and the horizontal

    tanx = vv / vh = 27.594 / 61.974

    x = 24.0*

    So, the velocity of the projectile 1.5s after firing is 67.8 ms-1 in the direction 24.0* above the horizon.

    Source(s): Myself~~~
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