PROJECTILE MOTION EQUATIONS EXAMPLES MOVIE

PROJECTILE MOTION EQUATIONS EXAMPLES MOVIE

How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Kinematic Equations: vxi = vicosθi, vyi = visinθi ,vxf = vxi xf = vxi t vyf = vyi – gt, yf = vyi t - (½)gt2, (vyf) 2 = (vyi)2 - 2gyf vxf = vxi = ? vyf = -gt xf = vxft, yf = - (½)gt2 Time to Bottom: t = √2y/(-g) = 3.19 s vx0 = (x/t) = 28.2 m/sĪ movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. Use the appropriate equations you may have to combine some of them.Ĥ Example 4.4: Non-Symmetric Projectile MotionĪ stone is thrown! xi = yi = 0 yf = m, vi = 20 m/s, θi = 30º a) Time to hit the ground? b) Speed just before it hits? c) Distance from the base of the building where it lands? Kinematic Equations vxi = vicosθi, vyi = visinθi vxf = vxi, xf = vxi t vyf = vyi - gt yf = vyi t - (½)gt2 (vyf) 2 = (vyi)2 - 2gyfĥ A stone is thrown! xi = yi = 0 vxi = vicosθi, vyi = visinθiĮxample 4.4: Solution A stone is thrown! xi = yi = 0 yf = m, vi = 20 m/s, θi = 30º a) Time to hit the ground? b) Speed just before it hits? c) Distance from the base of the building where it lands? First, calculate vxi = vi cos(θi) = 17.3 m/s vyi = vi sin(θi) = 10.0 m/s (Time when yf = m) yf = -45m = vyi t - (½)gt2 A general quadratic must be solved using the quadratic equation! This gives: t = 4.22 s Kinematic Equations vxi = vicosθi, vyi = visinθi vxf = vxi, xf = vxi t vyf = vyi - gt yf = vyi t - (½)gt2 (vyf) 2 = (vyi)2 - 2gyfĦ A stone is thrown! xi = yi = 0 vxi = vicosθi, vyi = visinθiĮxample 4.4: Solution A stone is thrown! xi = yi = 0 yf = m, vi = 20 m/s, θi = 30º a) Time to hit the ground? b) Speed just before it hits? c) Distance from the base of the building where it lands? First, calculate vxi = vi cos(θi) = 17.3 m/s vyi = vi sin(θi) = 10.0 m/s thit = 4.22 s b) Velocity just before it hits? vxf = vxi, vyf = vyi – gt so vxf = 17.3 m/s vyf = 10 – (9.8)(4.22) = m/s Speed (vf)2 = (vxf)2 + (vyf)2 vf = 35.8 m/s Angle: tan(θf) = (vyf/vxf) = -(31.3/17.3) = -1.8 θf = -60.9º Kinematic Equations vxi = vicosθi, vyi = visinθi vxf = vxi, xf = vxi t vyf = vyi - gt yf = vyi t - (½)gt2 (vyf) 2 = (vyi)2 - 2gyfħ A stone is thrown! xi = yi = 0 vxi = vicosθi, vyi = visinθiĮxample 4.4: Solution A stone is thrown! xi = yi = 0 yf = m, vi = 20 m/s, θi = 30º a) Time to hit the ground? b) Speed just before it hits? c) Distance from the base of the building where it lands? First, calculate vxi = vi cos(θi) = 17.3 m/s vyi = vi sin(θi) = 10.0 m/s thit = 4.22 s vf = 35.8 m/s, θf = -60.9º c) Distance from the base of th building where it lands? xf = vxi thit = (17.3)(4.22) = 73.0 m Kinematic Equations vxi = vicosθi, vyi = visinθi vxf = vxi, xf = vxi t vyf = vyi - gt yf = vyi t - (½)gt2 (vyf) 2 = (vyi)2 - 2gyfĨ vxi = vicosθi, vyi = visinθi ,vxf = vxi xf = vxi t, vyf = vyi – gtĮxample 4.2: The Long Jump A long-jumper leaves the ground at an angle θi = 20° above the horizontal at a speed of vi = 8.0 m/s a) How far does he jump in the horizontal direction? (Assume his motion is equivalent to that of a particle.) b) What is the maximum height reached? Kinematic Equations vxi = vicosθi, vyi = visinθi ,vxf = vxi xf = vxi t, vyf = vyi – gt yf = vyi t - (½)gt2 (vyf) 2 = (vyi)2 - 2gyfĪ long-jumper leaves the ground at an angle θi = 20° above the horizontal at a speed of vi = 8.0 m/s a) How far does he jump in the horizontal direction? (Assume his motion is equivalent to that of a particle.) b) What is the maximum height reached? vxi = vi cos(θi) = 7.5 m/s vyi = vi sin(θi) = 4.0 m/s a) How far does he jump in the horizontal direction? Range = R = (2vxivyi/g) = 2(7.5)(4)/(9.8) R = 7.94 m Kinematic Equations vxi = vicosθi, vyi = visinθi ,vxf = vxi xf = vxi t, vyf = vyi – gt yf = vyi t - (½)gt2 (vyf) 2 = (vyi)2 - 2gyfĪ long-jumper leaves the ground at an angle θi = 20° above the horizontal at a speed of vi = 8.0 m/s a) How far does he jump in the horizontal direction? (Assume his motion is equivalent to that of a particle.) b) What is the maximum height reached? vxi = vi cos(θi) = 7.5 m/s vyi = vi sin(θi) = 4.0 m/s R = 7.94 m b) What is the maximum height? h = h = 0.72 m Kinematic Equations vxi = vicosθi, vyi = visinθi ,vxf = vxi xf = vxi t, vyf = vyi – gt yf = vyi t - (½)gt2 (vyf) 2 = (vyi)2 - 2gyfĪ movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. Remember that vx never changes, & that vy = 0 at the highest point. Solve for the x and y motions separately. Decide on the time interval this is the same in both directions, & includes only the time the object is moving with constant acceleration g. Read the problem carefully, & choose the object(s) you are going to analyze.

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Projectile motion equations examples