How Long Does a Jumping Fish Take to Pass a Point Above Water?

[CaveatLector]

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 many times and cannot get the correct answer. Can someone please post a step by step solution so I can see where I am going wrong? Thanks in advance.

 

[*melinda*]

Can someone please post a step by step solution so I can see where I am going wrong?

We don't do that here.

However, if you show the work you've already done on this problem, people will point out your mistakes and help you solve your own problem.

 

[CaveatLector]

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 to solve for the time it takes to go all the way up to zero velocity and then solve again for time to get to .4m above the water and add that number to the previous up to zero velocity?

 

[Mindscrape]

Yes, the acceleration due to gravity will be negative if you use up as a positive reference. So, you'll get a quadratic with two answers, and pick the larger one.

 

[civil_dude]

You need to know that the direction of the acceleration is not positive, but negative if your velocity is positive.

Also, you might want remember that your equation should yield two answers and you have to know which one to use for the fish 'falling' past 0.4m

 

[*melinda*]

...do you first have to solve for the time it takes to go all the way up to zero velocity and then solve again for time to get to .4m above the water and add that number to the previous up to zero velocity?

I think you've got the right idea.

You want to split it into two problems; one for going up, and the other for coming back down.

 

[Mindscrape]

I think you've got the right idea.
You want to split it into two problems; one for going up, and the other for coming back down.

No, no. You only need to use the function for displacement.

 

[CaveatLector]

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?

 

[Mindscrape]

Did you get .819s?

Show us what you did.

 

[CaveatLector]

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

 

[CaveatLector]

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

 

[andrewchang]

you have enough information to use the equations

 

[CaveatLector]

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

 

[CaveatLector]

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

 

[CaveatLector]

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

 

[Mindscrape]

Huh? What's their answer?

 

[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, just needed to figure out the logistics of how to approach it. Thanks.