Problem number 25 from Chapter 4 in \"Physics forScientistsandEngineers\" (8th e

Question

Problem number 25 from Chapter 4 in "Physics forScientistsandEngineers" (8th ed.) - Serway|Jewett A boy stands on a diving board and tosses a stoneintoaswimming pool. The stone is thrown from a height of 2.50mabovethe water surface with a velocity of 4.00 m/s at theangleof60.0o above the horizontal. As the stonestrikesthewater surface, it immediately slows down to exactly halfthespeedit had when it struck the water and maintains that speedwhileinthe water. After the stone enters the water, it moves inastraightline in the direction of the velocity it had when itstruckthewater. If the pool is 3.00 m deep, how much timeelapsesbetweenwhen the stone is thrown and when it strikes thebottom ofthepool. I don't understand how to do this problem. I can setupadiagram and I know what it is looking for, but I don't knowhowtoget there. Problem number 25 from Chapter 4 in "Physics forScientistsandEngineers" (8th ed.) - Serway|Jewett A boy stands on a diving board and tosses a stoneintoaswimming pool. The stone is thrown from a height of 2.50mabovethe water surface with a velocity of 4.00 m/s at theangleof60.0o above the horizontal. As the stonestrikesthewater surface, it immediately slows down to exactly halfthespeedit had when it struck the water and maintains that speedwhileinthe water. After the stone enters the water, it moves inastraightline in the direction of the velocity it had when itstruckthewater. If the pool is 3.00 m deep, how much timeelapsesbetweenwhen the stone is thrown and when it strikes thebottom ofthepool. I don't understand how to do this problem. I can setupadiagram and I know what it is looking for, but I don't knowhowtoget there.

Explanation / Answer

The height from which the stone is thrown is h = 2.50 m The initial speed of the stone is u = 4.00 m/s The angle at which the stone is thrown is =60.0o The time of flight of the stone is t = (u * sin/g) The speed of the stone after it strikes the water surfaceis v = (u/2) = (4.00/2) = 2.00 m/s The height of the swimming pool is H = 3.00 m The time taken by the stone to travel inside the water andreach the bottom of the pool is t1 = (H/v) The time elapsed between when the stone is thrown and when itstrikes the bottom of the pool is T = t + t1 = (u * sin/g) + (H/v) The time of flight of the stone is t = (u * sin/g) The speed of the stone after it strikes the water surfaceis v = (u/2) = (4.00/2) = 2.00 m/s The height of the swimming pool is H = 3.00 m The time taken by the stone to travel inside the water andreach the bottom of the pool is t1 = (H/v) The time elapsed between when the stone is thrown and when itstrikes the bottom of the pool is T = t + t1 = (u * sin/g) + (H/v)

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