Lecture 7

Lecture 7Projectile MotionProjectile MotionOften studied in terms of rectangular components, since the projectiles acceleration always acts in the vertical direction.When air resistance is neglected, the only force acting on the projectile is its weight, which causes the projectile to have a constant downward acceleration of g = 9.81 m/s2 or 32.2 ft/s2.

Hence ay = -g and ax = 0.

VxVHorizontal MotionSince ax = 0, application of the constant acceleration equations yields

The 1st and last equations indicate that the horizontal component of velocity always remains constant during the motion. Vertical MotionSince the positive y axis is directed upward, then ay = -g

Note: Only two of the above three equations are independent of one another. Example 1A sack slides off the ramp, with a horizontal velocity of 12 m/s. If the height of the ramp is 6 m from the floor, determine the time needed for the sack to strike the floor and the range R where sacks begin to pile up.

SolutionCoordinate system. The origin of coordinates is established at the beginning of the path, point A. The initial velocity of a sack has components (VA)x =12 m/s and (VA)y = 0. Also, between points A and B the acceleration is ay = -9.81 m/s2. Since (VB)x = (VA)x = 12 m/s, the three unknowns are (VB)y, R and the time of flight t. The vertical distance from A to B is known, and therefore we can obtain a direct solution for t by using the equation.

Horizontal motion: Since t has been calculated, R is determined as follows:

Example 2The chipping machine is designed to eject wood chips at VO = 25 ft/s. If the tube is oriented at 30 from the horizontal, determine how high, h, the chips strike the pile if they land on the pile 20 ft from the tube.

Coordinate system: When the motion is analyzed between points O and A, the three unknowns are represented as the height h, the time of flight toa and the vertical components of velocity (VA)y . With the origin of coordinates at 0, the initial velocity of a chip has components ofSolution

Also, (VA)x = (VO)x = 21.65 ft/s and ay = -32.2 ft/s2. Since we do not need to determine (VA)y , we haveHorizontal motion

Vertical motion. Relating toa to the initial and final elevations of a chip, we have

Example 3Using a video camera, it is observed that when a ball is kicked from A, it just clears the top of a wall at B as it reaches its maximum height. Knowing that the distance from A to the wall is 20 m and the wall is 4 m high, determine the initial speed at which the ball was kicked. Neglect the size of the ball.

4 m 11SolutionCoordinate system. The three unknowns are represented by the initial speed VA , angle of inclination , and the time tAB to travel from A to B. At the highest point, B, the velocity (VB)y = 0, and (VB)x = (VA)x = VA cos .Horizontal Motion

Vertical Motion(1)

(2)(3)To obtain VA , eliminate tAB from Eqs 1 and 2, which yields

(4)Solve for in Eq 4. and substitute into Eq. 3, so that

Then using Eq. 4, the required initial speed is