3.2 Projectile Motion

Transcription

1 Motion in 2-D: Last class we were analyzing the distance in two-dimensional motion and revisited the concept of vectors, and unit-vector notation. We had our receiver run up the field then slant Northwest. We then determined how far the quarterback would need to throw the ball to get it to the receiver by breaking each vector into components So, we know what route the play calls for the wide receiver to run, we have determined how far the quarterback needs to throw the ball by using the following concepts: Recall that the displacement / distance was capable of being divided into two and three dimensions, such as: with the Magnitude of the displacement vector being: r = x 2 + y 2 + z 2 and the angle of the displacement vector being: θ = tan 1 ( y x)

2 However, one thing we have not analyzed up to this point is the motion of the ball. So the quarterback will throw the ball and it will take a path that goes up into the y-dimension and over in the x-dimension; then after meeting its maximum height, the ball will start coming back down in the y-dimension but keep going over in the x-dimension. As with the displacement vector, we can make problems simpler by breaking down the velocity vector into two and three dimensions. Written in unit vector notion so that: NOTE: Velocity Vector v is tangent to the particle s path at the particle s position. Also as we did with the displacement vector, where we knew the components of the vector, then we could determine the magnitude and direction (angle) of the vector. Velocity can be analyzed in the same way such as: NOTE: Velocity Vector v is tangent to the particle s path at the particle s position. Magnitude of Velocity Vector: v = v x 2 + v y 2 + v z 2 Angle of a Velocity Vector: tanθ = v y vx

3 As with displacement and velocity, we can greatly simplify problems by analyzing acceleration by its components along each axis. So if we write it in unit-vector notation, we get: Magnitude of Acceleration Vector: a = a x 2 + a y 2 + a z 2 Angle of an Acceleration Vector: tanθ = a y ax The general point of all the above information is that physics problems can be solved easier by analyzing the problem based on what dimension it is in. Thus, try to break problems into analyzing the motion of a ball on the x-axis, the y-axis, and the z-axis. This is a very important concept when dealing with projectile motion problems and force problems. (CONTINUE TO NEXT PAGE TO BEGIN DISCUSSING PROJECTILE MOTION)

4 Projectile Motion: Projectile Motion is when something is projected or launched. Essentially, it is an object that after being thrown, projected, or launched, exhibits no other acceleration or forces acting on it except for gravity (because you are high-school students, we will neglect air resistance). So when a baseball or football is thrown, a tennis ball hit, that is an example of projectile motion. An airplane or a duck is not an example of projectile motion. Before we get started, let s make the assumption that air has no effect on the projectile. A projectile is launched with an initial velocity v o that can be written as We can write the components of v ox and v oy can be found if we know the angle θ o : IMPORTANT: Important: During a particles two-dimensional motion, the projectile s position vector r and velocity vector v change continuously, but its acceleration vector a is constant and always directed vertically downward. The projectile has no horizontal acceleration.

5 - Therefore, we can break-up a problem of two-dimensional motion into its components: 1. Horizontal Motion with zero acceleration 2. Vertical Motion with constant downward acceleration NOTICE: The horizontal component of velocity vector remains constant, while the vertical velocity component changes continuously. Let s quickly analyze the motion of a ball being dropped versus a ball being projected vertically downward:

6 (HORIZONTAL MOTION): Because there is no acceleration in the horizontal direction, the horizontal component v x of the projectile s velocity remains unchanged from its initial value of v 0x throughout the motion. So previously when we were analyzing one-dimensional motion with no acceleration, we used this formula: x x o = v ox t Because we have v ox = v o cosθ o which makes the equation become (VERTICAL MOTION): x x o = (v ox cosθ o )t So vertical motion is exactly the same as it was when we threw a ball straight up and it came straight down. So basically, we just use the free-fall acceleration equations. v yf = v o sinθ o + gt v yf 2 = (v o sinθ o ) 2 + 2g (y y o ) y y 0 = (v o sinθ o )t gt2 y y 0 = v yf t 1 2 gt2 y y o = 1 2 (v yo + v yf )t

7 As when a ball is thrown vertically upward, it is directed upward initially and its magnitude steadily decreases to zero, which marks the maximum height of the path. Then ball changes direction and begins to start coming down. Note: Vertical motion and horizontal motion are connected through the variable of time. The time from the launch of the projectile to the time it hits the target is the same for both vertical motion and horizontal motion. Therefore, solving for time in one of the dimensions, vertical or horizontal, automatically gives you time for the other dimension. (TRAJECTORY) The entire motion of a projectile can be described by the parabolic formula (y = ax + bx 2 ) known as the Equation of the Path or Trajectory: The Trajectory equation can be found by combining the following equations: x x 0 = v 0 cos θ 0 t y y 0 = (v o sin θ 0 )t gt2 (HORIZONTAL RANGE): The horizontal range R of the projectile is the horizontal distance the projectile has traveled when it returns to its initial (launch) height. R = x x o = (v o cosθ o )t R = v o 2 g sin (2θ o)

8 Where sin(2θ o ) = 2 sin θ o cos θ o sin θ o cos θ o = 1 2 sin(2θ o) IMPORTANT: The horizontal range R is maximum for a launch angle of 45. THE ABOVE EQUATION DOES NOT GIVE THE HORIZONTAL DISTANCE TRAVELED BY A PROJECTILE WHEN THE FINAL HEIGHT IS NOT THE LAUNCH HEIGHT.