Question

A person in the passenger basket of a hot-air balloon throws a ball horizontally outward from the basket with speed 10 m/sm/s (Figure 1).

Figure

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Part A

Part complete

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/sm/s relative to the ground during this throw?

Part B

Part complete

Express your answer to three significant figures and include the appropriate units.

Part C

Part complete

What initial velocity (magnitude and direction) does the ball have relative to a person standing on the ground if the hot-air balloon is descending at 6.0 m/sm/s relative to the ground?

Answer 1

Solution :

Given :

v_x = 10 m/s

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❋ If the hot-air balloon is rising at 6.0 m/s :

Then : v_y = 6 m/s

Thus : Initial velocity of the ball relative to a person standing on the ground will be given by : v² = (v_x)² + (v_y)²

∴ v² = (10 m/s)² + (6 m/s)²

∴ v² = (10 m/s)² + (6 m/s)²

∴ v² = (136 m/s)²

⁂ v = 11.66 m/s

And, Direction : θ = tan^-1 {v_y / v_x} = tan^-1{ (6 m/s) / (10 m/s) } = 30.96^o (Above horizontal)

⁂ θ = 30.96^o (Above horizontal)

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❋ If the hot-air balloon is descending at 6.0 m/s :

Then : v_y = - 6 m/s

Thus : Initial velocity of the ball relative to a person standing on the ground will be given by : v² = (v_x)² + (v_y)²

∴ v² = (10 m/s)² + ( - 6 m/s)²

∴ v² = (10 m/s)² + ( - 6 m/s)²

∴ v² = (136 m/s)²

⁂ v = 11.66 m/s

And, Direction : θ = tan^-1 {v_y / v_x} = tan^-1{ ( - 6 m/s) / (10 m/s) } = - 30.96^o.

⁂ θ = 30.96^o (Below horizontal)

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