A person in the passenger basket of a hot-air balloon throws a ball horizontally outward from the basket with a speed of 10.0 m/s. What init

Question

A person in the passenger basket of a hot-air balloon throws a ball horizontally outward from the basket with a speed of 10.0 m/s. What initial velocity (magnitude and direction) does the ball have relative to a person standing on the ground if the hot-air balloon is rising at 6.0 m/s relative to the ground during this throw?

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Linh Đan 1 year 2021-09-03T20:58:32+00:00 1 Answers 178 views 0

Answers ( )

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    2021-09-03T21:00:24+00:00

    Answer:

    Explanation:

    Given

    balloon is rising with a speed of [tex]u_y=6\ m/s[/tex]

    Person throws a ball out of basket with a horizontal velocity of [tex]u_x=10\ m/s[/tex]

    Considering upward direction to be positive

    When ball is thrown it has two velocity i.e. in upward direction and in horizontal direction so net velocity is

    [tex]v_{net}=\sqrt{(u_x)^2+(u_y)^2}[/tex]

    [tex]v_{net}=\sqrt{(6)^2+(10)^2}[/tex]

    [tex]v_{net}=\sqrt{36+100}[/tex]

    [tex]v_{net}=\sqrt{136}[/tex]

    [tex]v_{net}=11.66\ m/s[/tex]

    Direction of velocity

    [tex]\tan \theta =\dfrac{u_y}{u_x}[/tex]

    [tex]\tan \theta =\dfrac{6}{10}[/tex]

    [tex]\theta =30.96^{\circ}[/tex]

    where [tex]\theta [/tex] is angle made by net velocity with horizontal .

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