- #1
revere21
- 10
- 0
Homework Statement
A cliff diver runs horizontally at 4.00 m/s. He hits the water 3.00 s later. Ignore air resistance.
(a) What is the diver's speed (magnitude of the velocity vector) just before he hits the water?
Homework Equations
Change in y-component: y(t) = vyot + (1/2)g(t^2)
Change in x-component: x(t) = vxot
V(t) = Vyo - gt
The Attempt at a Solution
To find the cliff height, I simply used -0.5(9.8)(3^2). So this comes out to 44.1 meters high.
To find the change in x-component (how far he traveled), I used x = v*t, which was (4.00)(3.00), which equals 12.0 meters. So far, super easy high school-level problem.
To find his speed, I plugged in 2.998 for time 1, and 2.999 for time 2 (two separate calculations, of course), so y(2.998) = vyot - 0.5(9.8)(2.998^2). Anyways, my answer keeps ending up as 29.4 m/s, which is one of the answers on the multiple choice test review (not a graded assignment, just a past test meant for review purposes), but the correct answer is 29.7 m/s. Regardless of how small of a time interval I calculate (i.e. if I calculate the difference between Velocity at time 2.9999998 seconds and 2.9999999 seconds) I cannot figure out where this 29.7 m/s comes from.
Please advise. Thanks everyone.