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Projectile Motion

Projectile Motion. Motion in 2 Dimensions. We know how to calculate motion on the x-axis…. In a race for the very last parking spot, a Valley senior came into the lot at a velocity of 20 m/s. If they slammed on their breaks and slid to a stop in 12 m… What was their acceleration ?

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Projectile Motion

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  1. Projectile Motion Motion in 2 Dimensions

  2. We know how to calculate motion on the x-axis… In a race for the very last parking spot, a Valley senior came into the lot at a velocity of 20 m/s. If they slammed on their breaks and slid to a stop in 12 m… • What was their acceleration? v2 =v02+2ax 0 m2/s2=400 m2/s2 + 2(12 m)a -400 m2/s2 = (24 m)a -16.67 m/s2 = a • How long did it take for them to come to a stop? v = v0+at 0 m/s = 22 m/s + (-16.67 m/s2)t -22 m/s = (-16.67 m/s2)t 1.32 s = t

  3. We know how to calculate motion on the y-axis… A very excited (and a little unfortunate) student, jumped from the top of the bleachers at the homecoming game in order to get to the concession stand. If the top of the bleachers is 12 m off the ground, and the student jumped up at a velocity of 3 m/s… • How fast is the student going when they landed on the ground below? v2 =v02+2ax v2 = 9 m2/s2 + 2(-9.8 m/s2)(-12 m) v2 = 9 m2/s2 + 470 m/s2 v = -16 m/s • How long does it take for them to get to the ground? v = v0+at -16 m/s= 3 m/s + (-9.8 m/s2)t -19 m/s = (-9.8 m/s2)t 1.9 s = t

  4. Now let’s put them together! A baseball pitcher throws a fastball at a speed of 42 m/s. It covers a distance of 18 m to cross home plate. x = 18 m v0 = 42 m/s v = 42 m/s This is what we said the motion looked like when we were studying 1 dimensional motion AND IT DOES… FROM ABOVE.

  5. But that’s only part of the story From the side, it looks more like this… We described the motion along the x-axis, but we didn’t take into account the motion in the y-axis. y = -0.9 m x = 18 m

  6. Along the x-axis… The ball moved with a nearly CONSTANT velocity of 42 m/s. (How do we know this?) This means that we can use the equation… vavg = Dd/Dt to find that the time is 0.43 s.

  7. Along the y-axis… • We know that the ball started with a velocity of 0 m/s • It accelerated towards the ground at -9.8 m/s2 • And (from the previous slide) the ball took 0.43 seconds to reach the batter. Using the formula y = v0t + ½ at2 We find that the total distance traveled on the y axis is -0.9 m.

  8. The different axes affect different attributes of an object’s motion Which part of the object’s motion (motion along the x-axis or motion along the y-axis)… • Determines how high the object will go? • Determines how far it will go? • Determines how long it will be in the air?

  9. Hangtime, Range, and Height These are the “science words” for • How long will the object be in the air? • How far will it go? And • How high will it go?

  10. Hangtime Hangtime is defined as the amount of time the projectile (object being launched) is traveling through the air. • It is determined by motion in the y-axis Gravityis what limits the projectile’s time in the air. • With every passing second, the object will accelerate towards the earth until it hits the ground. • The horizontal velocity, on the other hand, will remain relatively constant. • An object’s hangtime will be greatest when it is shot straight up at a 90o angle and is determined by the formula -2v0y/a = t Where a = g = -9.8 m/s2andt = hangtime

  11. Range Range is defined as the distance the projectile travels on the x-axis. • It is determined by the formula, vx = x/t , where t is the hangtime and x is the range An object’s range will be greatest when it is shot at a 45o angle. • The hangtimeis limited by gravity (y-axis) and is greatest when there is no motion on the x-axis (projectile is shot straight up). • However, the motion must have an x component in order for the object to have a non-zero range. • At 45o, the x and y components of the object’s velocity are balanced providing the object with the maximum possible combination of height and range.

  12. Height Height is defined as the distance the projectile travels on the y-axis. • It is determined by the formula, y= v0yt + ½ at2 • where t is the HALF of the hangtime, a is g (-9.8 m/s2), and y is the height. • An object’s height will be greatest when it is shot straight up at a 90o angleand will decrease as q approaches 0o. • At 45o, the x and y components of the object’s velocity are equal providing the object with the maximum possible combination of height and range.

  13. So, how do we figure out the different parts of the motion? Separate any given velocity at the angle it makes to the horizontal. vy = v sin q vx = v cosq v = 5 m/s • vy = v sin q • vy = (5 m/s) sin 37o • vy = 3 m/s • q = 37o • vx= v cosq • vx= (5 m/s) cos37o • vx= 4 m/s

  14. To Solve 2-D motion Problems • Determine the x and y components of the motion • List the x and y components in separate lists of “givens” • Solve by substituting givens into the appropriate formulas

  15. Vector Kinematic Equations x- equations vx= v cosθ vx = v0x + at x = ½ (v0x +vx) t x = v0xt + ½ axt2 vx2 = v0x2 + 2axx y- equations vy = v sin θ vy = v0y + at y = ½ (v0y +vy) t y = v0yt + ½ ayt2 vy2 = v0y2 + 2ayy

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