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What does it mean "It has slope -2/3 and x-intercept of 3"?
I sat around the other day thinking how I could apply vectors. Then I thought:
If I had a pistol and wanted to see how far the bullet went if I shot it, and my pistol could fire a bullet at 350 m/s, how far would it go?
Assumptions:
Air resistance is negligible. I am firing with my pistol parallel to the ground (or at right angles). The terrain I am shooting out to is FLAT surface. The acceleration due to gravity is 9.8 metres per second per second downwards. I am holding the pistol 1.6m above the ground.
Am I correct in saying it would travel 1.4km before hitting the ground 4 seconds after being fired?
Here's my working:
The acceleration vector of the bullet is -9.8j, which means the velocity is -9.8tj + c At time t = 0, v = 350i ==> 350i = 0j + c ==> c = 350i ==> v = 350i + (-9.8t)j ==> the position vector is 350ti + (-4.9t^2)j + C At time t = 0, position = 1.6j ==> 1.6j = C ==> p = 350ti + (-4.9t^2)j + 1.6j ==> p = 350ti + (1.6 - 4.9t^2)j Bullet hits the ground at vertical component = 0: 4.9t^2 = 1.6 ==> 49t^2 = 16 ==> t^2 = 16 ==> t = 4 So the bullet hits the ground after 4 seconds. At time t = 4, vertical component:
350 * 4 = 1400m
Yeah, kicking myself. To "KSmrq," it's not homework. I haven't done maths at school for two years. I guess you probably assumed so because the answer works out to be exactly 200m, which was purely coincidental. Also, I searched Google for the speed at which a standard 9mm pistol is fired. It told me 200-500m/s, so I took the mean. If I went out and performed this, with all those assumptions in mind, would it still be 4/7 of a second? —Preceding unsigned comment added by 138.217.32.202 ( talk • contribs)
Mathematics desk | ||
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< February 8 | << Jan | February | Mar >> | February 10 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
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The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
What does it mean "It has slope -2/3 and x-intercept of 3"?
I sat around the other day thinking how I could apply vectors. Then I thought:
If I had a pistol and wanted to see how far the bullet went if I shot it, and my pistol could fire a bullet at 350 m/s, how far would it go?
Assumptions:
Air resistance is negligible. I am firing with my pistol parallel to the ground (or at right angles). The terrain I am shooting out to is FLAT surface. The acceleration due to gravity is 9.8 metres per second per second downwards. I am holding the pistol 1.6m above the ground.
Am I correct in saying it would travel 1.4km before hitting the ground 4 seconds after being fired?
Here's my working:
The acceleration vector of the bullet is -9.8j, which means the velocity is -9.8tj + c At time t = 0, v = 350i ==> 350i = 0j + c ==> c = 350i ==> v = 350i + (-9.8t)j ==> the position vector is 350ti + (-4.9t^2)j + C At time t = 0, position = 1.6j ==> 1.6j = C ==> p = 350ti + (-4.9t^2)j + 1.6j ==> p = 350ti + (1.6 - 4.9t^2)j Bullet hits the ground at vertical component = 0: 4.9t^2 = 1.6 ==> 49t^2 = 16 ==> t^2 = 16 ==> t = 4 So the bullet hits the ground after 4 seconds. At time t = 4, vertical component:
350 * 4 = 1400m
Yeah, kicking myself. To "KSmrq," it's not homework. I haven't done maths at school for two years. I guess you probably assumed so because the answer works out to be exactly 200m, which was purely coincidental. Also, I searched Google for the speed at which a standard 9mm pistol is fired. It told me 200-500m/s, so I took the mean. If I went out and performed this, with all those assumptions in mind, would it still be 4/7 of a second? —Preceding unsigned comment added by 138.217.32.202 ( talk • contribs)