1. ## Projectiles help.

Im really stuck with this =(.
A girl stands at a waters edge and throws a stone from a height of 0.9meters. To get the stone to bounce, it must hit the water at an angle of 15 degrees. Find the minimum speed which she can throw the stone.
Well i calculated the time for the stone to fall...this is 0.429 seconds (its correct in the answers).
The vertical component of the stone hitting the water, the speed is 4.2m/s (although you may not need this, this was just worked out in an earlier part to question).
So i have that t = 0.429...but it's the only component i have of the horizontal...how do i work out the speed :S?

Thanks

2. You seem to be assuming that the girl throws the stone horizontally (so that the vertical velocity is initially zero). I'm not sure that the question is worded in such a way as to justify this assumption. (In practice, if you're trying to skim a flat stone on a lake, you probably tend to throw it in a slightly downwards direction, don't you?) But if the answer t=0.429 is given as correct, then that assumption must be correct. It certainly makes the problem a good deal easier. So let's assume that the vertical velocity is initially 0.

If the direction of travel is at an angle $\theta$ below the horizontal, then $\tan\theta = \frac{\text{vertical velocity}}{\text{horizontal velocity}}$. From that equation, knowing that the vertical velocity is 4.2m/s at the moment of impact, you can find the horizontal velocity at that moment, given that $\theta = 15^\circ$. And assuming that the horizontal velocity is constant, you have the answer to the question.

3. Hello, Ashley!

I think I have this solved . . .

A girl stands at a waters edge and throws a stone from a height of 0.9 m.
To get the stone to bounce, it must hit the water at an angle of 15°.
Find the minimum speed which she can throw the stone.

Well, i calculated the time for the stone to fall: 0.429 seconds (correct in the answers).
The vertical component of the stone hitting the water, the speed is 4.2 m/s.
Assuming she throws the stone horizontally, your calculations are correct.

As you found, the vertical component of the speed is: $4.2$ m/sec.

Let the original (horizontal) speed be $v$. .This is constant throughout the flight.

When the stone strikes the water, we have this diagram:
Code:
                  v
* - - - - - - - - - - - *
*  15°              |
*               |
*           | 4.2
*       |
*   |
*

In the right triangle: . $\tan15^o \:=\:\frac{4.2}{v} \quad\Rightarrow\quad v \:=\:\frac{4,2}{\tan15^o} \:=\:15.67461339$

Therefore: . $v \:\approx\:15.67$ m/sec.

4. Originally Posted by Opalg
You seem to be assuming that the girl throws the stone horizontally (so that the vertical velocity is initially zero). I'm not sure that the question is worded in such a way as to justify this assumption. (In practice, if you're trying to skim a flat stone on a lake, you probably tend to throw it in a slightly downwards direction, don't you?) But if the answer t=0.429 is given as correct, then that assumption must be correct. It certainly makes the problem a good deal easier. So let's assume that the vertical velocity is initially 0.

If the direction of travel is at an angle $\theta$ below the horizontal, then $\tan\theta = \frac{\text{vertical velocity}}{\text{horizontal velocity}}$. From that equation, knowing that the vertical velocity is 4.2m/s at the moment of impact, you can find the horizontal velocity at that moment, given that $\theta = 15^\circ$. And assuming that the horizontal velocity is constant, you have the answer to the question.
But if she throws it from a stationary position, wouldn't the initial velocity be 0?

Thank-you for your replies =). The problem is i have not been taught that equation lol. =(...Im going to look through my textbook now, you've given me a nice prod in the right direction =D thank-you!

EDIT: i realise the trig behind it now. Thank-you =)

5. But if she throws it from a stationary position, wouldn't the initial velocity be 0?
no. why do you think it would be?

6. Originally Posted by skeeter
no. why do you think it would be?
Because it goes from the hand and when its in the hand won't the velocity be 0 :S?

7. a common misconception from those not schooled in the details of classical mechanics. ... note that the object isn't a "projectile" until it leaves the hand.

8. Originally Posted by skeeter
a common misconception from those not schooled in the details of classical mechanics. ... note that the object isn't a "projectile" until it leaves the hand.
I didn't type out the question on full and i missed out the key word 'thrown horizontally' which others picked up on =), so vertical initial speed would be 0 right?
But thanks for your replies, it has helped me understand some stuff 'note that the object isn't a "projectile" until it leaves the hand.'
Cheers =)