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15 August, 01:20

A projectile is fired with an initial velocity of 450 feet per second at an angle of 70° with the horizontal.

Using the formulas of the projectile motion, in how many seconds will the projectile strike the ground? (Round your answer to the nearest tenth of a second.)

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  1. 15 August, 01:34
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    After 26.28 seconds projectile returns 26.28 seconds.

    Explanation:

    Initial velocity = 450 ft/s = 137.16 m/s

    Angle, θ = 70°

    Consider the vertical motion of projectile,

    When the projectile return to the ground we have

    Displacement, s = 0 m

    Acceleration, a = - 9.81 m/s²

    Initial velocity, u = 137.16 x sin70 = 128.89 m/s

    Substituting in s = ut + 0.5 at²

    s = ut + 0.5 at²

    0 = 128.89 x t + 0.5 x (-9.81) x t²

    t² - 26.28 t = 0

    t (t - 26.28) = 0

    t = 0 s or t = 26.28 s

    After 26.28 seconds projectile returns 26.28 seconds.
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