A plane, diving with constant speed at an angle of 44.1° with the vertical, releases a projectile at an altitude of 611 m. The projectile hi

Question

A plane, diving with constant speed at an angle of 44.1° with the vertical, releases a projectile at an altitude of 611 m. The projectile hits the ground 5.40 s after release. (a) What is the speed of the plane? (b) How far does the projectile travel horizontally during its flight? What were the magnitudes of the (c) horizontal and (d) vertical components of its velocity just before striking the ground? (State your answers to (c) and (d) as positive numbers.)

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Thành Đạt 7 days 2021-07-16T06:50:22+00:00 1 Answers 1 views 0

Answers ( )

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    2021-07-16T06:52:03+00:00

    Answer:

    Explanation:

    Let the velocity of projection be V .

    component of velocity in vertically downward direction = v cos 44.1

    component of velocity in horizontal direction = v sin 44.1

    s = ut + 1/2 gt²

    611 = v cos 44.1 x 5.4 + 1/2 x 9.8 x 5.4²

    = 611 = 3.88 v + 142.88

    v = 120.65 m /s

    b )

    horizontal component of velocity of projectile = v sin 44 .1 = 120.65 sin44.1

    = 83.96 m /s

    horizontal displacement = 83.96 x 5.4 = 453.38 m

    c ) horizontal component will remain unchanged so horizontal component

    =  v sin 44 .1 = 120.65 sin44.1

    = 83.96 m /s

    d ) velocity of projectile just before striking the ground

    v = u + gt

    = v cos 44.1  + 9.8 x 5.4

    =  120.65 xcos 44.1  + 9.8 x 5.4

    = 86.79 + 52.92

    = 139.71 m /s

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