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

Explanation:

Given

balloon is rising with a speed of $$u_y=6\ m/s$$

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

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

$$v_{net}=\sqrt{(u_x)^2+(u_y)^2}$$

$$v_{net}=\sqrt{(6)^2+(10)^2}$$

$$v_{net}=\sqrt{36+100}$$

$$v_{net}=\sqrt{136}$$

$$v_{net}=11.66\ m/s$$

Direction of velocity

$$\tan \theta =\dfrac{u_y}{u_x}$$

$$\tan \theta =\dfrac{6}{10}$$

$$\theta =30.96^{\circ}$$

where $$\theta$$ is angle made by net velocity with horizontal .