- #1

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- Homework Statement
- A canon located 60.0 m from the base of a vertical 25.0 m cliff, shoots a 15 kg shell at 43 degrees above the horizontal toward the cliff. What must be the minimum muzzle velocity, v1, for the shell to clear the top of the cliff?

- Relevant Equations
- v_2y = v_1y - gt

(v_2y)^2 = (v_1y)^2 - 2ghy

x = v_1x*t

I have attempted the problem a few times, and have posted my work below. I thought to check my work using

## v_x t = v_1 \times \cos\theta \times t = \Delta x##

to see if I get back expected results, and I did for the horizontal motion (the substitutions are from the numbered steps below):

## v_x t = v_1 \times \cos\theta \times t ##

## v_x t = \left(\frac{1}{cos\theta} \Delta x \sqrt{\frac{g}{2\Delta y}} \right) \times \cos\theta \times \left( \sqrt{\frac{2\Delta y}{g}}\right)##

## v_x t = \Delta x##

But when I check my work in the vertical direction, I get the curious ##\Delta y = \frac{\Delta x}{2}\tan\theta##. I expected to get back my ##\Delta y = 25.0## m (or very close to that) when I back substituted the numbers but I got ##\Delta y = 27.976## m for ##\Delta x = 60.0## and ##\theta = 43^0##. The difference of almost 3 is too much of a discrepancy for me to think things are correct. Can anyone tell me what I did wrong, or am I missing something when I check my work?

mass = ##15## kg

Get the time of flight from (0, 0) to (60, 25).

(1) ##v_{2y} = v_{1y} - gt##

(2) ##v_{1y} = gt## ; we only need to clear 25.0 m, so ##v_{2y} = 0.## (and I don't have a better idea)

(3) ##v_{2y}^2 = v_{1y}^2 - 2g\Delta y##

(4) ##v_{1y}^2 = 2g\Delta y## ; again we only need to clear 25.0 m, so ##v_{2y} = 0.##

(5) ##g^2t^2 = 2g\Delta y## ; sub (2) into (4)

(6) ##t = \sqrt{\frac{2\Delta y}{g}}## ; solve (5) for ##t##

Get the horizontal motion, ##v_x##; it is constant

(7) ##v_x t = \Delta x##

(8) ##v_x = \Delta x / t##

(9) ##v_x = \Delta x \sqrt{\frac{g}{2\Delta y}}## ; sub (6) into (8)

Get the vertical motion ##v_{1y}##

(10) ##\tan\theta = \frac{v_{1y}}{v_x} ##

(11) ##v_{1y} = \Delta x \sqrt{\frac{g}{2\Delta y}} \tan\theta## ; solve (10) for ##v_{1y}## and sub (9) in it

Now get ##v_{1}##

(12) ##v_{1}^2 = v_x^2 + v_{1y}^2 ##

(13) ##v_{1}^2 = \Delta x^2 \frac{g}{2\Delta y} + \Delta x^2 \frac{g}{2\Delta y} \tan^2\theta ## ; sub (9) and (11) into (12)

(14) ##v_{1}^2 = (1 + \tan^2\theta) \Delta x^2 \frac{g}{2\Delta y} ##

(14) ##v_{1} = \sec\theta \Delta x \sqrt{\frac{g}{2\Delta y}} ##

##v_1 = (\sec43^0)60\left(\sqrt{\frac{9.81}{50}}\right) \approx 36.339 m/s##

## v_x t = v_1 \times \cos\theta \times t = \Delta x##

to see if I get back expected results, and I did for the horizontal motion (the substitutions are from the numbered steps below):

## v_x t = v_1 \times \cos\theta \times t ##

## v_x t = \left(\frac{1}{cos\theta} \Delta x \sqrt{\frac{g}{2\Delta y}} \right) \times \cos\theta \times \left( \sqrt{\frac{2\Delta y}{g}}\right)##

## v_x t = \Delta x##

But when I check my work in the vertical direction, I get the curious ##\Delta y = \frac{\Delta x}{2}\tan\theta##. I expected to get back my ##\Delta y = 25.0## m (or very close to that) when I back substituted the numbers but I got ##\Delta y = 27.976## m for ##\Delta x = 60.0## and ##\theta = 43^0##. The difference of almost 3 is too much of a discrepancy for me to think things are correct. Can anyone tell me what I did wrong, or am I missing something when I check my work?

**given**: ##\Delta y = 25.0## m, ##\Delta x = 60.0## m, ##\theta = 43^0##,mass = ##15## kg

**find**: ##v_1##Get the time of flight from (0, 0) to (60, 25).

(1) ##v_{2y} = v_{1y} - gt##

(2) ##v_{1y} = gt## ; we only need to clear 25.0 m, so ##v_{2y} = 0.## (and I don't have a better idea)

(3) ##v_{2y}^2 = v_{1y}^2 - 2g\Delta y##

(4) ##v_{1y}^2 = 2g\Delta y## ; again we only need to clear 25.0 m, so ##v_{2y} = 0.##

(5) ##g^2t^2 = 2g\Delta y## ; sub (2) into (4)

(6) ##t = \sqrt{\frac{2\Delta y}{g}}## ; solve (5) for ##t##

Get the horizontal motion, ##v_x##; it is constant

(7) ##v_x t = \Delta x##

(8) ##v_x = \Delta x / t##

(9) ##v_x = \Delta x \sqrt{\frac{g}{2\Delta y}}## ; sub (6) into (8)

Get the vertical motion ##v_{1y}##

(10) ##\tan\theta = \frac{v_{1y}}{v_x} ##

(11) ##v_{1y} = \Delta x \sqrt{\frac{g}{2\Delta y}} \tan\theta## ; solve (10) for ##v_{1y}## and sub (9) in it

Now get ##v_{1}##

(12) ##v_{1}^2 = v_x^2 + v_{1y}^2 ##

(13) ##v_{1}^2 = \Delta x^2 \frac{g}{2\Delta y} + \Delta x^2 \frac{g}{2\Delta y} \tan^2\theta ## ; sub (9) and (11) into (12)

(14) ##v_{1}^2 = (1 + \tan^2\theta) \Delta x^2 \frac{g}{2\Delta y} ##

(14) ##v_{1} = \sec\theta \Delta x \sqrt{\frac{g}{2\Delta y}} ##

##v_1 = (\sec43^0)60\left(\sqrt{\frac{9.81}{50}}\right) \approx 36.339 m/s##