# 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 2021-09-03T21:06:50+00:00 1 Answers 37 views 0

The ball travels at a speed of 11.662 m/s, making a 30.96° angle from the horizontal ground.

Explanation:

If the ball is thrown horizontally with a speed of 10.0 m/s and the hot air balloon is rising at a speed of 6.0 m/s then we can make the following deductions:

1. The ball is ALSO rising at a speed of 6.0 m/s

2. The vertical speed of the ball is 6.0 m/s

Thus the ball travels with 10 m/s horizontal and 6 m/s vertical speed relative to the ground.

To find the total speed and direction:

Total speed = $$\sqrt{(V_X)^2+(V_Y)^2}$$

Total speed = $$\sqrt{10^2 + 6^2}$$

Total speed = 11.662 m/s

Angle from the horizontal:

Tan( Angle ) = Perpendicular / Base

Tan (Angle) = Vertical Speed / Horizontal Speed

Tan (Angle) = 6/10

Angle = 30.96°