Yahoo Answers: Answers and Comments for Projectile Motion Question Thanks!!!? [Physics]
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From unbertw07
enUS
Wed, 25 Feb 2009 17:21:36 +0000
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Yahoo Answers: Answers and Comments for Projectile Motion Question Thanks!!!? [Physics]
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From sapna: a million. First we would desire to discover h...
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Thu, 15 Dec 2016 23:37:42 +0000
a million. First we would desire to discover how long the cannonball is interior the air, its "cling time" in case you will. This span of time is the time it takes for the cannonball to bypass up, and fall bypass into opposite. it particularly is punctiliously based on the yfactor to the fee, so we would desire to continually use some trigonometry to discover this. in case you drew a suitable triangle, you will discover that the ycomp onent: velocity at y axis = (preliminary velocity)sin(seventy 9) = 993sin79 = 974.756 m/s. 2. Now which you have the fee on the y axis, we are able to now discover the time it takes for the cannon to land. we are able to apply the equation very final velocity = preliminary velocity + (Acceleration)(Time) or Vf = Vi + at word that your aceeleration = g, because of the fact this is pulling it downward so: Vf = Vi  gt. keeping apart t: Vf + gt = Vi gt = Vi  Vf t = (Vi  Vf)/g all of us understand the preliminary velocity interior the y axis, that's 974.756 m/s. yet what relating to the in simple terms suitable velocity? this is something thrilling: once you throw a ball up at say a definite velocity 2m/s, while it falls back on your hand (assuming you stored your hand on the comparable place) that is going to return on your hand at a velocity of 2m/s, the different of the preliminary velocity. So interior the case of the cannonball, i will say that the in simple terms suitable velocity is 974.756 m/s. i understand my Vi, Vf, and g, so i will now sparkling up for time. t = (974.756  (974.756))/9.8 = (974.756 + 974.756)/9.8 = 1949.512/9.8 = 198.9298 seconds. 3. Why do i'd desire to understand the cling time? that is because of the fact the ball is traveling on the x course for this plenty time, and after that factor, it stops shifting (except it keeps on rolling, yet this is yet another difficulty! enable's anticipate it maintains to be placed the place it lands.) besides, i will use the equation velocity = distance/time or distance = (velocity)(time) = vt I particularly have my time, yet i want my velocity. keep in mind we are searching for the gap on the xcourse, no longer the y, so we for the fee here, we would desire to get the xfactor to the projectile's velocity. utilising trigonometry lower back: velocity at x axis = (preliminary velocity)cos(seventy 9) = 993cos79 = 993(0.19) = 189.473 m/s. So, utilising this on the previous equation, and the airtime: distance traveled alongside the horizontal course = vt = (189.473)(198.9298) = 37,691.826 meters.

From Anonymous: The appropriate formula to describe the change...
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Wed, 25 Feb 2009 17:49:58 +0000
The appropriate formula to describe the change in position in the x direction is
Δx = v0_x*t + (1/2)gt^2
The acceleration acting on the x direction is 0,so
xf = v0_x*t
Converting this into polar coordinates you get
xf = (v0cos(Θ)t)
The time it takes for an object in a symmetrical trajectory to hit the ground can be determined from this equation: Δy = v0_y*t + (1/2)gt^2. Set it equal to 0 because you want to find the time it takes when it hits flat on the ground
0 = v0sin(Θ)t  (1/2)gt^2
gt^2 = 2v0sin(Θ)t
t =2 v0sin(Θ)/g
The distance it takes for an object to reach the other end of the trajectory is therefore
v0^2 sin(2Θ)/g = h
Solving for angle it is (1/2)arcsin(gh/v0^2)

From halligan: absolute zero.
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Wed, 25 Feb 2009 17:41:09 +0000
absolute zero.