A daring 510-N swimmer dives off a cliff with a running horizontal leap, What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is 1.75 wide and 9.00 below the top of the cliff
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Question “A daring 510-N swimmer dives off a cliff with a running horizontal leap, What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is 1.75 wide and 9.00 below the top of the cliff”
A daring 510-N swimmer dives off a cliff with a running horizontal leap, What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is 1.75 wide and 9.00 below the top of the cliff?
Answer
h= (1/2), g t2
18 = 9.8 t^2
t = 1.36 Falling
In 1.36 seconds, you can go horizontally 1.75m in 1.75 seconds
1.75 = u t = u (1.36)
u = 1.29 m/s horizontal speed
Conclusion
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