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- Projectile MotionObjectives:• Understand projectile motion and how gravity influences it• Understand the effects of projection speed, angle, and relative height on projectile motion• Learn to compute the maximum height, flight time, and flight distance of a projectile What is a Projectile?• A projectile is a body or object that – is in the air – is subject only to the forces of gravity and air resistance (i.e. the object is in free fall).• The motion of the center of mass of any object in free fall is governed by the laws of projectile motion 1
- Influence of Gravity • Gravity: pull of the mass of the Earth on a body • Gravity accelerates an object in a vertical direction towards the center of the Earth. • Acceleration due to gravity (g) is always straight downward at a constant 9.81 m/s2 (32.2 ft/s2). velocity velocity g = 9.81 m/s2 g = 9.81 m/s2 upward velocity downward velocity decreases increases Gravity & Vertical Velocity• From the laws of constant acceleration: (vvertical)2 = (vvertical)1 + (-9.81 m/s 2) * ∆t• Vertical velocity changes linearly with time vvertical decelerating at max. vhorizontal height vvertical (m/s) upward time (s) 0 accelerating g = 9.81 m/s2 downward 2
- Gravity & Horizontal Velocity• Gravity does not change the horizontal velocity of an object.• From the laws of constant acceleration: (vhorizontal)2 = (vhorizontal)1 + ahorizontal * ∆t For gravity, ahorizontal = 0, so: vvertical (vhorizontal)2 = (vhorizontal )1 vhorizontal vvertical changes g = 9.81 m/s2 vhorizontal remains constant Projectile Motion• Gravity causes a projectile to move in a parabolic path that is symmetric about the apex (the highest point in the trajectory) apex Height (m) Horizontal Distance (m) 3
- Influences on Projectile Trajectory• Three factors that influence projectile trajectory: – Angle of projection – Projection speed – Relative height of projection = (projection height) – (landing height) ed spe ion ject Pro Projection angle Projection height Influences of Projection Angle• Effect of projection angle on object trajectory (projection speed = 10 m/s, projection height = 0) 6 15 deg 5 30 deg 45 deg 4 Height (m) 60 deg 3 75 deg 90 deg 2 1 0 0 2 4 6 8 10 Distance (m)• Trajectory shape depends only on projection angle 4
- Influences of Projection Speed • Effect of projection speed on object trajectory (projection angle = 45°, projection height = 0) 5 2 m/s e gl 4 4 m/s an 6 m/s n tio Height (m) 3 ec 8 m/s oj 10 m/s Pr 2 1 0 0 2 4 6 8 10 Distance (m) Influences of Projection Height• Effect of relative projection height on object trajectory (projection speed = 10 m/s, projection angle = 45°) 5 (-2) m 4 (-1) m 0m Relative Height (m) 3 (+1) m 2 (+2) m 1 0 Landing Height 0 2 4 6 8 10 12 -1 -2 -3 Distance (m) Relative projection ht. = (projection ht.) – (landing ht.) 5
- Optimum Projection Conditions• Projection angle for maximum distance depends on relative projection height – rel. projection ht. > 0 optimal angle < 45° – rel. projection ht. = 0 optimal angle = 45° – rel. projection ht. < 0 optimal angle > 45° Rel. Proj. Optimal Max. Distance Height Angle Distance @ 45° +1 m 42.4° 11.15 m 11.11 m 0 45° 10.19 m 10.19 m -1 m 48.1° 9.14 m 9.07 m• Projection angle for maximum height = 90° Actual Projection Conditions• In real-life, often cannot attain theoretical optimum conditions• Trade-off exists between projection speed, angle, and height due to anatomical constraints Sport Actual Projection Angles Long Jump 18 – 27° Ski Jump 4 – 6° High Jump 40 – 48° Shot Put 36 – 37° (Hall, 2003) 6
- Trade-off Between Factors• Can obtain the same distance or height with different combinations of projection speed, angle, and height 6 5 Speed = 10.75 m/s Angle = 60° 4 Height (m) 3 2 Speed = 10 m/s 1 Angle = 45° 0 0 2 4 6 8 10 Horizontal Distance (m) Maximum Height• At the apex, vvertical = v0 sinθ = 0• From the laws of constant acceleration: v22 = v 12 + 2 a * d 0 = (v0 sinθ)2 + 2 (-9.81 m/s 2) * (yapex – y0) yapex = y 0 + (v0 sinθ)2 2 * (9.81 m/s 2 ) where: v0 θ yapex = height at apex y0 = projection height v0 = projection speed y0 θ = projection angle 7
- Example Problem #1 A high jumper leave the ground with a velocity of 6 m/s at a projection angle of 40°. Her center of mass is 1 m above the ground at take-off. What is the maximum height of her center of mass during the jump? Flight Time• From the laws of constant acceleration: d = v 1 * ∆ t + (½) a * ( ∆ t)2(yfinal – y0) = (v0 sinθ) * tF + (½)*(-9.81 m/s 2)* tF2 Solve the above quadratic equation to find the flight time tF (choose the largest positive answer) where: v0 yfinal = final height θ y0 = projection height v0 = projection speed y0 y final θ = projection angle 8
- Example Problem #2 A figure skater is attempting a jump in which she performs 3 complete revolutions while in the air. She leaves the ice with a velocity of 7 m/s at a projection angle of 30° If she spins at 3 revolutions per second, will she be able to complete all 3 revolutions before landing? Flight Distance• During projectile motion, vhorizontal = v0 cos θ is constant• From the laws of constant acceleration with a = 0: d = v1 * ∆t dF = (v 0 cos θ) * tF where: dF = flight distance v0 θ tF = flight time v0 = projection speed θ = projection angle dF 9
- Example Problem #3A kicker is attempting field goal from 40 yards away.The ball is kicked with an initial velocity of 24 m/s at a projection angle of 30°.The crossbar of the goal post is 10 ft above the ground.If his aim is correct, will he make the field goal? Effects on Projectile Motion Variable Determined by: Projection speed Horizontal velocity Projection angle Projection speed Vertical velocity Projection angle Vertical velocity Maximum height Projection height Vertical velocity Flight time Projection height Final height Horizontal velocity Flight distance Flight time 10
- Influence of Air Resistance• In real-life, air resistance will cause both horizontal and velocity to change while in flight.• Forces created by wind will also affect the trajectory No Air Resist Air Resist Tailwind Height (m) Oblique Trajectory Horizontal Distance (m) 11

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