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Understanding Projectile Motion in 2-Dimensional Space

Explore the principles of projectile motion in a 2-dimensional space, with examples and equations to calculate height, velocity, and distance covered. Understand the concept of acceleration and independence of horizontal and vertical motion.

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Understanding Projectile Motion in 2-Dimensional Space

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  1. (a) 1, (b) 3

  2. Fig. 4-6, shows the direction of acceleration for a particle located at its tail, and its length (representing the acceleration magnitude) can be drawn to any scale.

  3. Acceleration is second derivative of time, therefore 1 and 3 ax and ay both are constant. hence a is constant. 2 and 4 ay is constant but ax is not, therefore a is not constant.

  4. A special case of two-dimensional motion: A projectile is an object upon which the only force acting is gravity.  Yes, see fig 4.9 Figure 4-9

  5. The horizontal motion and the vertical motion are independent. Figure 4-10 One ball is released from rest at the same instant that another ball is shot horizontally to the right. Their vertical motions are identical. Eq. 2-15 Eq. 2-11 Eq. 2-16

  6. horizontal distance Solving (21) for x-xo=R, substituting y-yo=0 into (22) will give; ; Since Solving (21) for t, substituting it into (22) will give; Therefore Vo=

  7. Examples of Projectile Motion: In the given picture below, Alice throws the ball to the +X direction with an initial velocity 10m/s. Time elapsed during the motion is 5s, calculate the height that object is thrown and Vy component of the velocity after it hits the ground.  -h=-1/2gt2

  8. John kicks the ball and ball does projectile motion with an angle of 53º to horizontal. Its initial velocity is 10 m/s, find the maximum height it can reach, horizontal displacement and total time required for this motion. (sin53º=0.8 and cos53º=0.6)

  9. In the given picture you see the motion path of cannonball. Find the maximum height it can reach, horizontal distance it covers and total time from the given information. (The angle between cannonball and horizontal is 53º and sin53º=0, 8 and cos53º=0, 6)

  10. Here xo and Ɵo = 0, to find t use;

  11. y v x = -2m, v = - (4m/s)j a = (4)2/2=16/2=(8m/s)i y = 2 v = - (4m/s)i a = - (8m/s)j a x a v

  12. The position of P with respect to A = Position of P with respect of B + position of B with respect to A.

  13. or sin70o or cos70o

  14. 0

  15. Ɵo=0 1.52 m

  16. N SE E SE Vpg = Vpw + Vwg

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