Finding Initial Speed Given Range

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To determine the initial speed required for a catapult to achieve a maximum range of 0.67 km, the relevant formula is Rmax = V0^2/g, where g is the acceleration due to gravity (9.8 m/s²). The discussion emphasizes the importance of knowing the launch angle, which should be 45 degrees for maximum range, simplifying the equation to V0^2/g. Participants express frustration with rote memorization of formulas, advocating for understanding the derivation of equations through velocity-time graphs. This approach is suggested as a more effective way to tackle various projectile motion problems. Mastering these concepts enhances problem-solving skills in ballistics and related fields.
reigner617
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Homework Statement


If you want to use a catapult to throw rocks, and the max range is 0.67 km, what initial speed must the rocks have as they leave the catapult?

Homework Equations



v=Δr/Δt

The Attempt at a Solution


I sketched a graph of the projectile trajectory with the desired range on the x-axis. I also converted 0.67 km to 670 m. I concluded that acceleration would be -9.8 m/s. I couldn't go any further since I feel like I am not given enough information to work with.
 
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I feel like I am not given enough information to work with.
... you need to know the angle the stones are launched at - you are told this though. The way the question is written suggests you have already derived or had given to you a bunch of equations for range, max-height and so on. If not, then try sketching velocity-time graphs for the horizontal and vertical components of the motion.

note: the acceleration is -9.8m/s/s in the +y direction, taking +y to be "upwards".
 
Thank you for the quick response. I used the formula Rmax=V02/g. After substituting the givens into the formula, I came up with 81 m/s
 
Did you understand where the equation came from?
 
Not quite, although a similar formula that I know is R=(V0^2)sin(2t)/g
 
I also understand that for the range to be max, the angle has to be 45, and if we put 45 into the V0^2sin(2t)/g formula, sin2t would just be 1, and we are left with just V0^2/g
 
reigner617 said:
Not quite, although a similar formula that I know is R=(V02)sin(2θ)/g ... I also understand that for the range to be max, the angle has to be 45, and if we put 45 into the V02sin(2t)/g formula, sin2θ would just be 1, and we are left with just V02/g
... well done :)

I just don't like the style of teaching that has students memorize a bunch of equations.
Note: that equation only works where the initial and final heights are the same - when you can derive the equations from velocity-time graphs you'll be able to do any ballistics problem, and many more besides, without having to memorize or look up any equations ... so it's a skill worth obtaining.
 

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