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Launching of a Potato (Involve kinetic and potential energy and an angle ) 
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#1
Oct1111, 08:02 PM

P: 4

A potato of mass 210 g is launched from a potato gun, being released from a height of 0.6 m, at an angle of 47o with respect to the horizontal, and with an initial speed of 38 m/s. Assume the effect of air resistance can be neglected.
mass = 210 g or 0.210 kg height (initial) = 0.6 m θ (angle) = 47 degrees Vo (initial velocity) = 38 m/s a. What is the horizontal component of the potato’s velocity at the time it is launched? b. What is the speed of the potato when it is its maximum height? Hint: the answer is not 0! c. Calculate the kinetic energy of the potato when it is launched. d. Calculate the gravitational potential energy of the potato when it is launched. e. Calculate the kinetic energy of the potato when it reaches its maximum height. f. Using the principle of energy conservation (that is, a Unit 2 technique not a Unit 1 technique), determine how high above the ground the potato rises before beginning its descent. Vectors have always been an issue for me in this class...as soon as I see an angle I get all mixed up. Part A: I'm not really sure what is meant by "horizontal component". The potato is being launched upward so it has an x and y value. Would its vertical component be 38 m/s in the y direction? I have no clue... Part B: This is projectile motion...the first thing I thought of is 0! Why is it not 0? Is it because the potato is still moving horizontally, just not vertically? Part C: Okay so K = 0.5 * m * v^2 K = 0.5 * 0.210 kg * 38 m/s ^2 K = 150 J Now there is an angle involved...I have no clue how or if that needs to be factored in... Part D: U = mgh U = 0.210 kg * 9.8 m/s^2 * 0.6 m U = 1.23 J And again with the angle... Part E: I think once I get the angle part dealt with I'll be able to solve the rest of this... Part F: I know you guys don't want to do our homework for us, which I completely understand, but I really can't do this problem correctly without knowing how to handle the angle. Once that part is pointed out, I should be able to do the rest of the problem. Thanks 1 million times to who ever can help!! =] 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 


#2
Oct1211, 04:32 AM

P: 950

In lower classes of Physics students only work problems where motion is in a straight line. But then in higher classes there will be problems where motion will be in a curve.
But the beautiful thing about projectiles projected under gravity is to notice that the curved motion is just made up of a) motion in a straight line with constant velocity along the horizontal and b) motion in a straight line with constant acceleration along the vertical. The particle is subjected simultaneously to these two types of motion and thus goes in a curve. The student can continue to work in straight line motion as he did in lower classes. But he has to be careful whether he is working in the horizontal or in the vertical since the conditions are different. 


#3
Oct1211, 08:27 AM

P: 6

Spoiler
A) You can seperate the velocity into two vectors, lets say v_{x} and v_{y}. The find the horizontal velocity, you simply make a triangle with the v_{initial} as the hypotenuse with an angle of 47 degrees. Use v_{initial}sin(x) to find v_{x} (the horizontal velocity). All this is assuming there is no air friction.
Spoiler
The speed of the potato I interpret as the magnitude of its velocity. Since v_{x} is 0m/s at max height we can neglect that and use v_{y}. The answer would be v_{i}sin(x), the same as your question A.
Spoiler
Use KE=.5mv^{2}.
Spoiler
Use PE=mgh. Keep in mind that h has a nonzero value because you are not on the ground.
Spoiler
KE=0.
Gotta go to class! My answers might not necessarily be correct, please check them over! I'm just a student myself lol. 


#4
Oct1211, 10:26 AM

Mentor
P: 11,813

Launching of a Potato (Involve kinetic and potential energy and an angle )
Consider the two figures here:
The first shows the trajectory of your projectile plotted on XY axes. The initial launch speed is indicated as V_{o} and the initial launch angle is θ. Note that a quantity with both magnitude and direction is a vector quantity. So we can call V_{o} with its angle the initial velocity vector. The second drawing shows the velocity vector drawn on a set of axes that represent the vertical (y) and horizontal (x) components of a velocity. What are the projected lengths of this vector on the axes? That is, what are V_{x} and V_{y}? You'll want to use a little trigonometry. Something to note is that while the velocity in the ydirection changes due to the acceleration due to gravity, the velocity in the xdirection is constant throughout the trajectory. 


#5
Oct1211, 12:32 PM

P: 950

[QUOTE ... find the horizontal velocity, you simply make a triangle with the v_{initial} as the hypotenuse with an angle of 47 degrees. Use v_{initial}sin(x) to find v_{x} (the horizontal velocity)... QUOTE]
If x is 47deg then sin must be replaced by cos. 


#6
Oct1211, 12:42 PM

P: 950

[QUOTE ... Since v_{x} is 0m/s at max height we can neglect that and use v_{y}[/QUOTE]
May I correct some typos: If v[itex]_{x}[/itex] is meant to be the horizontal component of the velocity, then v_{x} must be replaced by v_{y} and v_{y} by v_{x}. 


#7
Oct1311, 02:02 AM

P: 6

Thanks for making the corrections :x



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