Recent content by picc84

Ok. For #3, for the first equation I got deltaX = deltaT (55 m/s) For the second equation, I got deltaX = deltaT (3.5 m/s squared) For the equation expressing the condition that the cycle catches up with the car, I got deltaX = Vi deltaT + 1/2(3.5)deltaT (squared) I got that the cycle...

I honestly don't know. Can I?

Aw, your right.

Hmm. Ok, after doing it again I got 18.81 meters for the depth of the hole. It took 0.06 seconds for the sound to get to her ear, so it actually took 1.94 seconds for the pebble to hit the water. Putting that into the equation I got 18.81 for the depth.

I actually came up with some answers to these questions but I want to be sure I have the *right* ones before I turn them in. If I have it right i'd appreciate the say-so if anyone minds the bother, but if I'm wrong maybe just some insight as to how to do better. Problem: A spelunker finds a...

Ok so the sun moved 15 degrees in the hour that passed. Thats 15 degrees period since it was theoretically at 0 degrees the hour before, which it must have been to cast no shadow at all. And i know the shadow is 3 ft long. Where do I go from here? gnpatterson, No, I don't know trig. I'm...

Directly overhead, I think. Which happens at noon, right? I don't know! I made a sketch of the situation like you said. If the sun was directly overhead, as it needs to be for no shadow, at noon, then one hour later it was one o'clock. Thats about all I've deduced. I think its the...

1) A student is standing on the east side of a building and notices that it casts no shadow. One hour later, she notices that the shadow of the building is about 3 feet long. Approx. how high is the building she is standing next to? In this problem I honestly just don't know where to start...