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Projectile Motion

Projectile Motion. Projectiles at an Angle. Projectiles at an Angle. Last lecture, we discussed projectiles launched horizontally. Horizontal projectiles are just one type of projectile problem that we can discuss. They are also one of the easiest. Projectiles at an Angle.

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Projectile Motion

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  1. Projectile Motion Projectiles at an Angle

  2. Projectiles at an Angle • Last lecture, we discussed projectiles launched horizontally. • Horizontal projectiles are just one type of projectile problem that we can discuss. • They are also one of the easiest

  3. Projectiles at an Angle • Now we will discuss projectiles that are launched at an angle other than 0°. • They still share many of the same characteristics of horizontal projectiles

  4. Projectiles at an Angle • As the animation to the right shows, horizontal and vertical movement are still independent of each other. • If we neglect air resistance, any projectile launched from a moving object will land in the same place it was launched if the moving object maintains the same speed.

  5. Projectiles at an Angle • Lets analyze the following problem and attempt to solve it. • A football is kicked with a velocity of 25 m/s at an angle of 60°. Calculate the range, total time in the air, and maximum height of the football’s path.

  6. Solving Problems • We need to start just like we did with horizontal projectiles and make an x and a y column. • We need to determine what part of the velocity of the ball is in the horizontal and what part is in the vertical. • We do this by finding the x and y components of the 25 m/s velocity at the angle of 60°. • By using trigonometry, we find the following: • Vx = Cos60(25m/s) =12.50 m/s • Vy = Sin60(25m/s) = 21.65 m/s

  7. X – column Vi = 12.50 m/s t = 4.42 sec d = vt d = (12.50)(4.42) d = 55.25m Y – column Vi = 21.65 m/s a = -9.8 m/s Vf = 0 (at max height) Vf = Vo + at 0 = 21.65 + (-9.8)(t) t = -21.65/-9.8 t = 2.21 seconds Vf2 = Vi2 + 2ad 0 = (21.652) + (2)(-9.8)(d) d = -468.72/-19.6 d = 23.91 m Solving Problems

  8. Important Considerations • One important consideration is that we must make the acceleration of gravity NEGATIVE in these calculations. • The reason has to do with the fact that we have an object moving both up and down. • We must use up as a positive value and down as a negative value.

  9. Important Considerations • Another assumption we must make is that at the maximum height, Vf in the y direction will be 0. • Understand that at the exact instant that the ball reaches the maximum height, it has no velocity up or down in the vertical direction.

  10. Important Considerations • Notice that when we calculate time in the y column, we are only finding how long it takes to reach the maximum height. • Time must be multiplied by 2 to determine the length of the entire trip

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