Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
A diver springs upward from a board that is 3.00 m above the water. At the instant she contacts the water her speed is 8.90 m/s...Asked by katherine
A diver springs upward from a board that is 2.98 meters above the water. At the instant she contacts the water her speed is 9.39 m/s and her body makes an angle of 70.4° with respect to the horizontal surface of the water. Determine her initial velocity.
Answers
Answered by
drwls
Her vertical velocity component when she hits the water is
Vy' = 9.39 sin70.4
Her horizontal velocity component is
Vx = 9.39 cos70.4
The latter component remains constant during the dive.
For the initial vertical velocity component, use
Vyo^2/2 + g*2.98 m = Vy'^2/2
Use the initial horizontal and vertical velocity components, Vx and Vyo, ffor the initial velocity.
Vy' = 9.39 sin70.4
Her horizontal velocity component is
Vx = 9.39 cos70.4
The latter component remains constant during the dive.
For the initial vertical velocity component, use
Vyo^2/2 + g*2.98 m = Vy'^2/2
Use the initial horizontal and vertical velocity components, Vx and Vyo, ffor the initial velocity.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.