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A cannon is placed at the bottom of a cliff 85 m high. If the cannon is fired straight upward, the cannonball just the reaches the top of the cliff.

a) Calculate the initial speed of the cannonball. b) Suppose a second cannon is placed at the top of the cliff and fired horizontally with the same initial speed as part (a). Prove numerically that the range of this cannon is the same as the maximum range of the cannon from the base of the cliff.

My work for part A work How do I know what the velocity of the ball is at the top of the cliff?

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Working on this now. Also, the velocity of the ball at the top of the cliff is zero because if it were not, it would go higher than 85m; the question tells you it just reaches the top. – Dylan Sabulsky Feb 12 '13 at 7:11
Hi user1530249 - could you please edit any work you need to show directly into the question? Image links shouldn't be used to display text. – David Z Feb 12 '13 at 10:26
This site uses MathJax to render nice looking equations so you don't have to attach an image. See here for a guide to typesetting math. It's pretty simple when you get the hang of a few basics. You have to place your equations between \$ \$ for MathJax to recognize them, or \$\$ \$\$ for block equations. – Michael Brown Feb 12 '13 at 13:35
1. lol typesetting is such a waste of time since he's already written the stuff 2.Hi dssfdfsd. Welcome to Physics.SE. This site deals with conceptual Physics Q&A. We don't encourage homework questions that doesn't involve any sort of work done by the author (which is you) and asks other users to solve the problem. If you think you could clarify your question, add what you've done along with your question. We're ready to help you. If you aren't clear, Please have a look at our homework policy for more info. After improving the post, flag it for moderator attention. – raindrop Feb 17 '13 at 17:27

For part A, the velocity at the top of the cliff is $v=0$ because it 'just reaches' the top of the cliff.

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Hello this is the my solution.

$V_y^2=2gh$ $\implies V_y=40.82$ $\text{m/s}$

and then for second cannon $V_x=V_y=40.82$ $\text{m/s}$. I assumed second cannon only has horizontal velocity.

Range of second cannon ball : $L = V_x \times t$ and we know $h=(gt^2)/2$

Found $t= 4.16$ $s$

then $L=4.16 \times 40.82=170.01$ $m$

Actually I dont know if I have understood right. Because final answer is not proving the ranges are same, indeed it shows that the ranges are different.That is because although times are same for both cannon ball previous one cannon ball has negative acceleration (gravitational). Second one it just moves with constant horizontal speed. If I understand something wrong let me know :)

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Your answer for a) seems fine. I personally would do this the following way: Newtons Laws give us this eqn of motion $$v_{f}^{2}=v_{i}^{2}+2ad \rightarrow ~=v_{i}^{2}-2*g*85$$ Solving this gives $40.837 \frac{m}{s}$.

For part b), I leave it to you to show the following. Using one of the eqns of motion, calculate how long it would take the ball to fall from 85m if it were merely dropped straight down. This is the time it will take for the ball to fall to earth when fired horizontally. Plug the time from the dead drop into your horizontal distance eqn and obtain the distance. Then calculate the cannon firing at 45 degrees with respect to the ground; you'll find what youre looking for.

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