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7-2 Projectile Motion. Independence of Motion in 2-D Projectile is an object that has been given an intial thrust (ignore air resistance)  Football,

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Presentation on theme: "7-2 Projectile Motion. Independence of Motion in 2-D Projectile is an object that has been given an intial thrust (ignore air resistance)  Football,"— Presentation transcript:

1 7-2 Projectile Motion

2 Independence of Motion in 2-D Projectile is an object that has been given an intial thrust (ignore air resistance)  Football, Bullet, Baseball Moves through the air under the force of gravity Path is called Trajectory  If you know the force of the thrust, you can determine trajectory

3 Independence of Motion in 2-D If you hit a golf ball, throw a football, only force acting on the projectile is gravity (long range) Gravity acts same in vertical direction Horizontal force has no effect on vertical component

4 Combination of vertical drop and horizontal moment give object a parabola trajectory

5 Strategy Separate into vertical and horizontal motion problem Vertical motion is treated like a straight up or down movement (g) Horizontal motion treated like a constant velocity problem  No thrust and air drag neglected  No horizontal forces acting (a = 0) Motions are connected by time variable  Solve for time in one of the dimensions and will give you other

6 Equations Y-direction: v y = -gt  y = y 0 – (1/2)gt 2  t = √ -2(y – y 0 ) g X-Direction  x = x 0 + v x0 t

7 Stone is thrown horizontally at 15 m/s from top a cliff 44 meters high A) How far from the base of the cliff does the stone hit the ground? Known  X 0 = 0v x0 = 15 m/s y 0 = 0v y0 = 0a = -g Unknown  X when y = -44 m  V at that time a F g = F net

8 How far from the base of the cliff does the stone hit the ground? Y-direction  y = y0 – (1/2)gt2  t = √ -2(y-y0)so: = √ -2(y) / g g  √ -2(-44 ) / -9.8 m/s2  = 3.0 sec

9 How far from the base of the cliff does the stone hit the ground? X direction  x = x 0 + v x0 t  x = (15 m/s)(3.0 m/s)  = 45 meters from the base

10 How fast is it moving the instant before it hits the ground? V y = -gtv x = -(9.80 m/s2)(3.0 s) = -29 m/s v y v v =√ v x 2 + v y 2 = √(15 m/s) 2 + (-29 m/s) 2 = 33 m/s

11 Projectiles Launched at an Angle Initial velocity has an initial horizontal and vertical component  Rises with slowing speed and falls with gaining speed Max Height  height of projectile when vertical velocity is zero and only has horizontal component  y = y i + V yi t – (1/2)gt 2 Range (R)  Horizontal distance the projectile travels

12 Problem A ball is launched with an initial velocity of 4.47 m/s at an angle of 66 o above the horizontal.  A) What is the max height the object attained?  B) How long did it take the ball to return to the launching height?  C) What was the range? Known  x i = 0y i = 0 v i = 4.47 m/sθ = 66 o a = -g Unknown  y when Vy = 0t = ??x when y = 0

13 Equations Needed Y direction:  V yi = vi sinθ(4.47)sin 66 o  V yi = 4.08 m/s  *** V y = V yi – gt  *** y = y i + V yi t – (1/2)gt 2 x direction  V xi = v i cos θ  V x = V xi  x = x i + v xi t

14 What is the max height the object attained? V y = 0, t = v yi / g t = (4.08 m/s) / (9.80 m/s 2 ) t = s y max = v yi t – (1/2)gt 2 = (4.08 m/s)(0.420) – (1/2)(9.80)(0.420) 2 y max = m

15 How long did it take the ball to return to the launching height? y = 0 y = y i + V yi t – (1/2)gt 2 0 = 0 + V yi t – (1/2)gt 2 t = 2v yi / g = 2(4.08 m/s) / (9.80) t = 0.83

16 What was the range? x = R R = V xi t = (4.47 m/s)(cos 66 o )(0.83 s) R = 1.5 m


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