 # 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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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, 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

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

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

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

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

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

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

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

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

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

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

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

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 = 0.420 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 = 0.850 m

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

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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