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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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2021-09-03T20:58:32+00:00
2021-09-03T20:58:32+00:00 1 Answers
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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 .