Projectile Motion Question Thanks!!!?

Projectile motion to date has been seen as how far something will travel at a given initial trajectory and velocity. The practical application of this is slightly different: few want to know about how far something will go. Rather, most applications involve targetting an object at some distance, given a particular initial velocity.

Question - Generate the well-formed, generic solution for determining angle from distance and initial velocity.


3 Answers

  • 1 decade ago
    Favorite Answer

    absolute zero.

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  • Anonymous
    1 decade ago

    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)

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  • sapna
    Lv 4
    3 years ago

    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 y-factor 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 y-comp 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 x-course, no longer the y, so we for the fee here, we would desire to get the x-factor 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 air-time: distance traveled alongside the horizontal course = vt = (189.473)(198.9298) = 37,691.826 meters.

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