Projectile Motion Problem - Diving

[Whotto]

This is a projectile motion problem at a Grade 12 level in Canada.

My main problem here is conceptual, not the math itself.

1. The given question

A diver takes off with a speed of 8.0m/s from a 3.0-m high diving board at 30^o above the horizontal. How much later does she strike the water?

My diagram:
[PLAIN]http://img830.imageshack.us/img830/3949/diagram.jpg

2. The solution given

I was given the solution to use the equation d = v₀t + 0.5at² and solve for t, which would end up in the form of the Quadratic formula. The answer given is 1.3s.

[PLAIN]http://img294.imageshack.us/img294/1028/worky.jpg

3. What I don't understand

I don't understand why the solution would work, or more precisely, be complete. From how I understand it, the solution would give the answer for just the red highlighted part in the projectile trail of this picture:

[PLAIN]http://img638.imageshack.us/img638/1486/diagram2b.jpg

Since the equation had distance (Or is it displacement?) = 3m, I think that it did not include anything above the diving board. Am I wrong here?

----------------------Alternative---------------------------

[PLAIN]http://img205.imageshack.us/img205/1645/diagram3.jpg

If I were given this situation, would I still arrive at the same answer with the same process as this? [PLAIN]http://img294.imageshack.us/img294/1028/worky.jpg

I hope I have described my problem well enough, any help would be appreciated!

Thanks!

 

[rock.freak667]

From d=v_ot-1/2gt²

They are finding when d=-3 (They have the origin at 3m high, so he hits the ground at 3m below the line)

-3=v_ot-1/2gt²

or

-1/2gt²+v_ot+3=0

 

[Whotto]

From d=v_ot-1/2gt²

They are finding when d=-3 (They have the origin at 3m high, so he hits the ground at 3m below the line)

-3=v_ot-1/2gt²

or

-1/2gt²+v_ot+3=0

Thanks for the quick reply rock freak!

So then d = displacement I take it. If the diving board was right on top of the water then d = 0?

 

[rock.freak667]

Thanks for the quick reply rock freak!

So then d = displacement I take it. If the diving board was right on top of the water then d = 0?

The formula is this actually:

d=d₀+v₀t+1/2at²

If you pick your origin on the top of the water then d₀=initial displacement from the origin = 3 (the diving board is fixed 3m into the air).

If you pick the origin where diving board is, then d₀=0, as the displacement from the origin is 0.

 

[Whotto]

The formula is this actually:

d=d₀+v₀t+1/2at²

If you pick your origin on the top of the water then d₀=initial displacement from the origin = 3 (the diving board is fixed 3m into the air).

If you pick the origin where diving board is, then d₀=0, as the displacement from the origin is 0.

I get what you've said so far, but I don't understand how it works into my problem.

I will try ask my question in another way:

Previous: [PLAIN]http://img638.imageshack.us/img638/1486/diagram2b.jpg
New situation: [PLAIN]http://img205.imageshack.us/img205/1645/diagram3.jpg

If I were given the new situation rather than the previous, would I still arrive at the same answer with the same process as I would with the previous? Why or why not?

 

[collinsmark]

If I were given the new situation rather than the previous, would I still arrive at the same answer with the same process as I would with the previous? Why or why not?

Hello Whatto,

No you would get a different answer. The difference is is the sign of the initial velocity. (Also, you might wish to double check your '+' and '-' signs in your quadratic. Remember, things relating to the up direction get a '+' sign, and things in the down direction get a '-' sign.)

So in the previous case we have (as has been pretty-much established):

0 = \frac{1}{2} \ at^2 +v_0 t - d

t = \frac{-4 \pm \sqrt{4^2 - 4 \left[ \left(\frac{1}{2} \right) (-9.8) \right] (3)}}{(2) \left[ \left( \frac{1}{2} \right)(-9.8) \right]}

Now try it again, with the new situation where the diver dives down instead of up. In other words, instead of

v_0 = 4 \left[ \frac{\mathrm{m}}{\mathrm{s}}\right]

Make it,

v_0 = -4 \left[ \frac{\mathrm{m}}{\mathrm{s}}\right]

 

[rl.bhat]

For the time being forget about the projectile.

If you through a stone from a height h with the velocity v in the downward direction and then in the upward direction, which stone takes a longer time to reach the ground and why?

 

[Whotto]

Hi Collinsmark,

It is much clearer to me now. I got 0.474 seconds with -4m/s instead.

So the equation 0 = \frac {1}{2}at^{2} + v_{0}t - d would include the height that the diver has gained while taken off, even though d = 3m anyways?

Thank you for the good explanation!

For the time being forget about the projectile.

If you through a stone from a height h with the velocity v in the downward direction and then in the upward direction, which stone takes a longer time to reach the ground and why?

Thanks for the reply!

I understand that part, and that is why I am getting confused on

So the equation 0 = \frac {1}{2}at^{2} + v_{0}t - d would include the height that the diver has gained while taken off, even though d = 3m anyways?

 

[collinsmark]

So the equation 0 = \frac {1}{2}at^{2} + v_{0}t - d would include the height that the diver has gained while taken off, even though d = 3m anyways?

The height that the diver is relative to the board (represented by d) varies with time t. You have set you your equations such that you are solving for the specific time t when d = -3.0 m.

In other words, d is not always -3.0 meters. But it is -3.0 meters at one specific point in time. You are solving for that point in time.

(And by the way, keep track of your '+' and '-' signs. The way the equations were originally set up with distances being relative to the board, d = -3.0 m, since the water is below the board. [See rock.freak667's explanation for setting up the equations differently if you wanted to -- you'll get the same answer either way though])

 

[rl.bhat]

In the kinematic equation the variables are vo, vi, a, d and t. In these variables t is always positive. The other variables may have the same direction or opposite direction. In a give problem variables having the same direction may be taken as positive and the remaining can be takes as negative. Now try applying this concept in the above problem and see what you get.

 

[Whotto]

Ah, I get it now!

Much appreciated and thank you for taking the time, rockfreak, collinsmark, and bhat!

and yes - I will definitely keep better track of my signs.