King Fahd University of Petroleum Minerals Mechanical Engineering

King Fahd University of Petroleum & Minerals Mechanical Engineering Dynamics ME 201 BY Dr. Meyassar N. Al-Haddad Lecture # 4

12. 6 Motion of a Projectile • Projectile: any body that is given an initial velocity and then follows a path determined by the effects of gravitational acceleration and air resistance. • Trajectory – path followed by a projectile

Horizontal Motion is Uniform Motion Notice that the Horizontal motion is in no way affected by the Vertical motion.

Projectile

Horizontal and vertical components of velocity are independent. Vertical velocity decreases at a constant rate due to the influence of gravity.

Verify this mathematically

Horizontal Motion • Acceleration : ax= 0 • Conclusion # 1: Horizontal velocity remains constant • Conclusion # 2: Equal distance covered in equal time intervals

Vertical Motion • ac= -g = 9. 81 m/s 2 = 32. 2 ft/s 2 • Conclusion # 1: Equal increments of speed gained in equal increments of time • Distance increases in each time interval

Projectile Motion • Assumptions: (1) free-fall acceleration (2) neglect air resistance • Choosing the y direction as positive upward: ax = 0; ay = - g (a constant) y • Take x 0= y 0 = 0 at t = 0 • Initial velocity v 0 makes an angle 0 with the horizontal v 0 x

Maximum Height At the peak of its trajectory, vy = 0. From Time t 1 to reach the peak Substituting into:

Projection Angle • The optimal angle of projection is dependent on the goal of the activity. • For maximal height the optimal angle is 90 o. • For maximal distance the optimal angle is 45 o.

Projection angle = 10 degrees

Projection angle = 45 degrees 10 degrees 30 degrees 45 degrees

Projection angle = 60 degrees 10 degrees 30 degrees 45 degrees 60 degrees

Projection angle = 75 degrees 10 degrees 30 degrees 45 degrees 60 degrees 75 degrees So angle that maximizes Range ( optimal) = 45 degrees

Example • A ball is given an initial velocity of V 0 = 37 m/s at an angle of = 53. 1. Find the position of the ball, and the magnitude and direction of its velocity, when t = 2. 00 s. Find the time when the ball reaches the highest point of its flight, and find its height h at this point • The initial velocity of the ball has components: v 0 x = v 0 cos θ 0 = (37. 0 m/s) cos 53. 1° = 22. 2 m/s v 0 y = v 0 sin θ 0 = (37. 0 m/s) sin 53. 1°= 29. 6 m/s a) position x = v 0 xt = (22. 2 m/s)(2. 00 s) = 44. 4 m y = v 0 yt - ½gt 2 = (29. 6 m/s)(2. 00 s) –½ (9. 80 m/s 2)(2. 00 s)2 = 39. 6 m

Solution (con. ) • Velocity • vx = v 0 x = 22. 2 m/s • vy = v 0 y – gt = 29. 6 m/s – (9. 80 m/s 2)(2. 00 s) = 10. 0 m/s

Solution (cont. ) • b) Find the time when the ball reaches the highest point of its flight, and find its height H at this point.

Solution (cont. ) c) Find the horizontal range R, (that is, the horizontal distance from the starting point to the point at which the ball hits the ground. )

A ball traveling at 25 m/s drive off of the edge of a cliff 50 m high. Where do they land? 25 m/s Initial Conditions vx = 25 m/s vy 0 = 0 m/ a =- 9. 8 s m/ 2 s t=0 Horizontally x = x 0 + (v 0)x t x = 25 *3. 19 = 79. 8 m Vertically v = v 0 -gt y = y 0 + v 0 t + 1/2 gt 2 …. -50 = 0+0+1/2(-9. 8)t 2 … t = 3. 19 s v 2 = v 02 - 2 g(y-y 0)…. y 0 = 0 m y =- 50 m x 0 =0 m 79. 8 m

Review • Example 12. 11 • Example 12. 12 • Example 12. 13