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 m wide and 9.00 m below the top of the cliff?
2007-02-20 17:22:21 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
If I did the math right... 1.29 m/s horizontally (assuming the leap adds no vertical initial velocity).
the answer is found by...
9m = .5 (9.8) * (t^2)
t^2= 9/(.5 * 9.8)
t= sqrt[9/(.5 * 9.8)]
velocityhorizontal = 1.75 / time
I believe Z-prime's answer is wrong. btw. Her time of drop is off.
1.83s would give a fall distance around.... 15-17 meters (approx, didnt do the math for 1.83 secs)
Also the 510 Newtons does not play into the problem, it is there to try to mislead you if you did not understand how to do the problem.
2007-02-20 17:44:15 · answer #1 · answered by Nick R 2 · 1⤊ 0⤋
Calculate the time for an object to fall 9 m from rest. Now calculate speed = 1.75m / t
You should have a formula in your textbook to solve for t.
2007-02-21 01:40:53 · answer #2 · answered by Roy E 4 · 0⤊ 0⤋
Well, it will take her SQR (2s/a) secs to fall 9m. a=9.81m/s. s=9, so it will take her 1.83 secs to fall. to clear a 1.75m wide ledge. In that time, you need to travel horizontally at least as fast as 1.75/1.83=0.95m/s. This isn't very fast. Even I can run that fast.
2007-02-21 01:37:37 · answer #3 · answered by zee_prime 6 · 0⤊ 0⤋