A football is thrown at an angle of 30.° above the horizontal. The magnitude of the horizontal component of the ball’s initial v

Question

A football is thrown at an angle of 30.° above the horizontal. The magnitude of the horizontal

component of the ball’s initial velocity is 13.0 meters per second. The magnitude of the vertical

component of the ball’s initial velocity is 7.5 meters per second. [Neglect friction.]

The football is caught at the same height from which it is thrown. Calculate the total time the football

was in the air. [Show all work, including the equation and substitution with units.]

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Dâu 4 years 2021-08-10T12:52:10+00:00 2 Answers 4 views 0

Answers ( )

  1. Cherry
    0
    2021-08-10T12:53:51+00:00
    Reply

    Answer:

    Explanation:

    Let the initial velocity is u.

    ux = 13 m/s  

    uy = 7.5 m/s

    Angle of projection, θ = 30°

    ux = u Cos 30 = 13

    u = 15 m/s

    Height caught is same

    use second equation of motion

    h = uy t – 0.5 x gt²

    0 = 75. t – 0.5 x 9.8 x t²

    t = 1.24 s

  2. minhkhue
    0
    2021-08-10T12:54:06+00:00
    Reply

    Answer:

    1.53 s

    Explanation:

    Initially vertical component of velocity of the ball, uy = 7.5 m/s

    Net displacement is vertical direction is zero, Δy =0

    Use second equation of motion:

    Δy = uy t + 0.5 a t²

    Here, acceleration a = -g                           (g =9.8 m/s²)

    Substitute all the values and solve for g

    0 = 7.5 t -0.5 (9.8)t²

    7.5 t = 4.9 t²

    t = 1.53 s

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