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A guy posted this problem on a forum:

There is a bird sitting on a pole of height h. you throw a rock at it and the moment the rock leaves your hand the bird starts flying horizontally away from you at 10m/s, the rock passes the point where the bird was when you threw it and then reaches twice the height of the pole. on its descent it touches the bird. what is the horizontal velocity of your rock.

Promise it's not homework. How would you solve it? This is what I've done so far:

Let's say were a distance $d$ away from the pole, and that the bird travels distance $x$ before we hit it. Then if $v_h$ is the horizontal velocity, we know that $d + x = v_ht_h$, and that $x = 10t_h$.

The problem is a bit vague but we can assume that $2h$ is the vertical peak of the projectile. Then, we have that $2h = v_vt_{up} - (1/2)gt_{up}^2$. The rock will travel down $h$ before it hits the bird, so another equation is $h = (1/2)gt_{down}^2$. Finally we have that $t_{up} + t_{down} = t_h$.

So $t_{down} = \sqrt{h/(4.9)}$. I'm not sure how to solve for $t_{up}$, especially because we don't know that the vertical component of velocity is. Any help would be appreciated!

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Hi helpworld. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. – Qmechanic May 5 '13 at 17:33
Is the answer 15 ? – Saurabh Raje May 7 '13 at 8:20
If it is not, I have gone wrong. Else, confirm, and then I will show you how I got it. – Saurabh Raje May 7 '13 at 8:23

You do know what the vertical component of the velocity is, because you know how high it got. You can do this in two obvious ways, but I prefer energy conservation. As you threw the rock, it's energy was entirely kinetic: $E_0 = \frac{1}{2} m v_v^2 + \frac{1}{2} m v_h^2$. At its peak, the energy was partly potential, with a kinetic contribution just from $v_h$: $E_\mathrm{p} = m g (2 h)+ \frac{1}{2} m v_h^2$. Energy conservation tells us that these are equal, so $v_v = \sqrt{2 g (2 h)} = 2\sqrt{gh}$.

Using this information, you'll be able to solve for the amount of time $t_1$ it took the rock to go up to $h$ the first time, and the amount of time $t_2$ it took to peak and come back down to $h$. Now, $t_2$ is the amount of time the bird had to fly before getting "touched" by the rock. Since you know its velocity, you know the horizontal distance between the top of the pole and the contact point. Of course, ($t_2-t_1$) is the amount of time the rock had to travel between the top of the pole and the bird. You now know the horizontal distance and the time, so you can figure out the horizontal velocity.

I won't fill in all the details since you seem to be happy solving the rest on your own. If you're still stuck, just ask.

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Given the total height requirement you can find the vertical velocity as $v_y = \sqrt{ 2 (2 h) g}$. Then the time to reach the height $h$ is first $t=\sqrt{\frac{h}{g} (2-\sqrt{2})}$ and then on the way back down $t=\sqrt{\frac{h}{g} (2+\sqrt{2})}$.

If the (horizontal) distance between you and the pole is $d$ then for the projectile to pass the pole at the first time above then $d = v_x \sqrt{\frac{h}{g} (2-\sqrt{2})}$

So the (horizontal) distance of the bird and you is

$$ x_B = d + v_B t = v_x \sqrt{\frac{h}{g} (2-\sqrt{2})} + v_B t$$

where $v_B=10$ is the bird velocity. To equate the above with the projectile position of $x_P = v_x t$ at time $t=\sqrt{\frac{h}{g} (2+\sqrt{2})}$ gives the soluton

$$ v_x = v_B \left(\frac{1}{2}+\frac{1}{\sqrt{2}}\right) $$

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