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14 April, 09:10

A bullet is fired at an angle of 60 degrees with an initial velocity of 200.0 m/s. How long is the bullet in the air? What is the maximum height reached by the bullet?

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  1. 14 April, 09:28
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    Time of flight, t = 35.34 sec.

    Maximum heght, H = 1530.60 m.

    Explanation:

    initial velocity of the bullet u = 200m/s

    Angle of projection θ = 60°

    (a).

    As we know that in projectile motion;

    Time of flight t = [2u*sin (θ) ] / g

    t = 2*200*sin (60°) / 9.8

    t = 35.34 sec.

    (b).

    As we khow that in projectile motion;

    Maximum heght, H = [u²*sin²θ]/2g

    = (200) ² * sin² (60) / 2*9.8

    Maximum heght, H = 1530.60 m.
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