What is the minimum speed needed to clear a tank of sharks while water skiing?

[notagenius08]

Homework Statement

A water skier is going to glide up a 2.0m frictionless ramp then sail over a 5.0m wide tank filled with sharks, she'll drop the tow rope at the base of the ramp. What minimum speed must she have as she reaches the ramp in order for her to clear the sharks.

Homework Equations

1/2mv^2=mgh
Vf^2=Vi^2 +2ax
t=sqrt of 2(height)/g

The Attempt at a Solution

figured out time to fall 2 meters = .63 seconds. So she'll need to cross the 5.0m wide tank in this amount of time. 5/.63 = 7.82 m/s this is the speed needed as she leaves the top of the ramp.

1/2mv^2=mgh solved for v
v=sqrt of 2gh
this is the speed needed to make it from the the bottom of the ramp to the top.
=6.26 m/s

add these two speed together and it should be velocity at bottom of ramp 6.26+7.82 =14.1 m/s

or so I thought. Book gives answer as 10 m/s.

so which calculation is wrong? 10-7.82 =2.18 m/s this isn't enough to make it up the ramp.

10-6.26 = 3.74 this isn't enough speed to make it over the sharks.

All I can think is that maybe I am calculating the speed to get up the ramp incorrectly.

Any help?

Thanks.

 

[Doc Al]

A water skier is going to glide up a 2.0m frictionless ramp

What's the angle of the ramp? I assume 2.0m is the height of the top of the ramp?

 

[notagenius08]

No angle given. In the drawring it is not linear, but curved like a slightly flattened 's', or a local min concave up, inflection point, then concave down local max, if that makes sense.

Yes 2.0m is the top of the ramp.

 

[Doc Al]

No angle given. In the drawring it is not linear, but curved like a slightly flattened 's', or a local min concave up, inflection point, then concave down local max, if that makes sense.

Yes 2.0m is the top of the ramp.

All that matters is the angle she moves when she leaves the ramp. We assume it's horizontal. (And that gives the correct answer.)

figured out time to fall 2 meters = .63 seconds. So she'll need to cross the 5.0m wide tank in this amount of time. 5/.63 = 7.82 m/s this is the speed needed as she leaves the top of the ramp.

Good!

1/2mv^2=mgh solved for v
v=sqrt of 2gh
this is the speed needed to make it from the the bottom of the ramp to the top.
=6.26 m/s

That's the speed she'd need to just make it up the ramp, with zero speed at the top.

add these two speed together and it should be velocity at bottom of ramp 6.26+7.82 =14.1 m/s

Ah... speeds don't add, but energy does.

Use conservation of energy from start to finish. She starts off with some unknown KE and ends up with PE + KE.

 

[notagenius08]

I always have trouble setting up the before and after, with conservation...

OK, PE+KE=KE so solving for Vi gives

Vi=sqrt of Vf^2 -2gh

but it should be

Vi=sqrt of Vf^2+2gh

But yes the second equation gives the right answer. Thank you very much.

 

[notagenius08]

Ok, nevermind, it is because I had Vi and Vf on the wrong sides of the equation, it does work out to be the second equation!

Thanks again.