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| Photo used with permission under Creative Commons license from Dennis Jarvis. |
Suppose a cannonball is fired vertically from ground-height at an initial velocity of 30 m/s. How long is the cannonball in flight before it crashes back to the ground? (Assume no air resistance in this problem.)The easy way to do this problem is calculate how long the cannonball takes to get to the peak of its trajectory:
v_f = v_i + a*t
0 = 30 + (-9.8) * t
t = 3.06 seconds
Since, by our symmetry argument, the cannonball spends an equal time going down as it does going up, we can double our result to find the overall time:
total time = 2 * (time going up)
total time = 2 * 3.06
total time = 6.12 seconds
While this is a quick and easy way to calculate the answer, does it make assumptions that are not true? Is the symmetry argument really valid in this case? Is there a way to determine the answer of 6.12 seconds of total flight time without using the symmetry argument?
In the comments, explain how it is possible to get t = 6.12 seconds without using the symmetry argument. What equations would you use and how do they work out?
In order to get an answer of 6.12s for the total flight time without using the symmetry argument, i first used the big d equation in order to find the peak height reached. a=(-9.8) V_i=30m/s V_f=0m/s d_i=0m t=3.06s to form an equation of d_f=0+30×(3.06)+1/2(-9.8)(3.06)^2 30×3.06 is 91.8 which you would then add to 1/2(-9.8)(3.06^2) which is -45.88 when you add the two together you get d_f=45.918 i then took that number, and used it as d_i for the trip back down, where v_i=0 a=(-9.8) d_i=45.918 d_f=0 i plugged those values into the big d equation, giving me 0=45.918+0(t)+1/2(-9.8)(t)^2 the zero times t cancels out, and after subtracting 45.918 from both sides and multiplying 1/2 and -9.8 your left with -45.918=-4.9t^2 after dividing both sides by -4.9, you get 9.37=t^2 or √9.37=t. The √9.37 is 3.06, which you can then add to the time the cannonball took to reach its peak (3.06s) giving you at total flight time of 6.12s.
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DeleteIn order to get an answer of 6.12s for the total flight time without using the symmetry argument, i first used the big d equation in order to find the peak height reached. a=(-9.8)m/s^2 V_i=30m/s V_f=0m/s d_i=0m t=3.06s to form an equation of d_f=0+30×(3.06)+1/2(-9.8)(3.06)^2. 30×3.06 is 91.8 which you would then add to 1/2(-9.8)(3.06^2) which is -45.88, when you add the two together you get d_f=45.918 i then took that number, and used it as d_i for the trip back down, where v_i=0m/s a=(-9.8)m/s^2 d_i=45.918m d_f=0m i plugged those values into the big d equation, giving me 0=45.918+0(t)+1/2(-9.8)(t)^2 the 0(t)+ is 0 and after subtracting 45.918 from both sides and multiplying 1/2 by -9.8 you are left with -45.918=-4.9t^2 after dividing both sides by -4.9, you get 9.37=t^2 or √9.37=t. The √9.37 is 3.06, which you can then add to the time the cannonball took to reach its peak (3.06s) giving you at total flight time of 6.12s.
In order two get 6.12 Sec as the total flight time without using the symmetry argument.First find the average velocity V_avg = (V_i+V_f)/2 Fill in values V_i=30,V_f=0 and (30+0)/2 = 15 m/s.
ReplyDeleteThe time is 3.06 seconds to reach the peak.
Using the formula D = R*T you simply fill in the values to find the distance to the peak. R= 15 m/s T= 3.06 sec
D= 15*3.06 = 45.9 meters.
Then using the little D equation d_f=1/2 * a * t^2 fill in values d_f=45.9,a=9.8 (down is the positive direction)
45.9=.5*9.8*t^2
45.9/4.9=t^2
t^2=9.36734693877551
√t^2=√9.36734693877551
T= 3.06
From the peak to the cannon or ground it takes the cannonball 3.06 seconds. This is the same time it takes from the cannon to the peak. Add the 3.06's together and 6.12sec is the total flight time of the cannonball.
Indeed, there is a way to determine the flight time of the cannonball was 6.12 seconds wiyhout having to use the symmetry argument. The first step you need to take is to look at the information that you were given. d_i=0 m since the cannonball started from the ground, v_i=30 m/s since that was given in the problem, a=-9.8 m/s^2 since that is the acceleration of gravity when an object is in free fall, as is described in this problem, the v_f of when the cannonball reaches the apex of its flight is 0 m/s since the cannonball has no velocity for a brief moment in time. Thetime for the upward flight of the cannonball is unknown so you would use the equation of v_f=v_i+at to solve for time.
ReplyDelete0=30+(-9.8)t you would subtract 30 from both sides
-30=(-9.8)t you would divide by-9.8
3.06=t is what you get for the time that it takes the cannonball to go upward. To find the time that it takes the cannonball to return back to earth is to find the distance the cannonball traveled by using the formula d=v_avg(t). But first you must find the v_avg by using the formula v_avg=(v_i+v_f)/2
v_avg=(30+0)/2
v_avg=15 then to find the total distance traveled you plug that value into the small distance formula
d=v_avg(t)
d=15(3.06)
d=45.9
This gives.you the distance the ball travels upward, which id the same distance it travels back to the earth's surface. You then plug the distance into the simplified.big d formula
d_f=.5(a)(t^2)
-45.9=(.5)(-9.8)(t^2)
-45.9=(-4.9)(t^2) divide.by (-4.9)
9.36735=(t^2) take the square root of both sides and since time.cannot.be negative you.get the answer of
3.06=t
you add both of the values of t together to receive the total timeof.flight of the cannonball which is equal to 6.12 seconds, which proves that the symmetry argument is.valid in this case.
Nice work this week!
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