Recent content by CaveatLector

Nevermind. I figured it out. Use velocity formula to find time at zero v. Use that time to solve for distance traveled at 4.5m/s over the t of .459. Then do the freefall formula to solve for time at .4m then add that time to previous t solved for. It worked. Makes sense intellectually...

a=4.905 or -4.905? b= 4.5 c= .4 plug those into quadratic eq.?

I'll give you a dollar if I can have your brain.

Well what am I doing wrong? Is it a 2 part problem?

Seems like I would need more info to solve this problem than what I have been given in the problem. Too many variables?

Yes that is what I got but it's wrong. So I'm doing something wrong here.

Using the formula I posted above I still get the wrong answer. I just plug the numbers in and solve for t right? I'm a little confused as to whether I need to approach the problem differently. Do I need to do 2 separate calculations, one for the the fish going up and then one for it coming down?

Okay the equation i used was y-yo=vot+(1/2)at^2. yo is zero because the fish starts at the water. y is the distance that the problem gives you which is .4m. a is 9.81, which is gravity. vo is 4.5m/s. But I am stuck. Do you use -9.81 since the fish is falling or do you first have...

A fish is able to jump vertically out of the water with a speed of 4.5 m/s. How much time does it take for the fish to pass a point 0.4 m above the water on the way down? Can someone please post how to do this? I know you need to solve for t by using the quad. equation but I have done it...