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Projectile Motion

Projectile Motion. Mathematical Application. Simultaneous Equations. Example: If the two equations above are simultaneous equations, then their common variable ( ) must be the same for both equations.

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Projectile Motion

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  1. Projectile Motion Mathematical Application

  2. Simultaneous Equations Example: If the two equations above are simultaneous equations, then their common variable () must be the same for both equations. Note: the number of unknown variables must equal to or less than the number of equations in order to solve.

  3. Substitution Method • choose one of the variables common for both equations • manipulate one of the equations to solve for that variable • combine the two equations by substituting the manipulated equality for the same variable in the second equation. • Solve the combined equation for the second unknown variable Given: c = 5 Unknown:a = ? b = ? Equations: Part I: -3 -3 Substitution: Solution: Part II: (Substitution Method)

  4. Example 1 A stone is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high. How long does it take the stone to reach the bottom of the cliff? What is Step 1? Draw a picture Δdx= ? vx = 5 m/s Δ tx= ? vi = 0 m/s g = -9.8 m/s2 Δdy= 78.4 m Δty= ? ΔtP= Δty= Δtx vf = ?

  5. 2Δd a 2 (78.4 m) (-9.8m/s2) t = Example 1 (continued) • Horizontal Vertical A stone is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high. How long does it take the stone to reach the bottom of the cliff? Δdy= 78.4 m vi = 0 m/s a = -9.8 m/s2 Vx = 5.0 m/s • G: • U: • E: • S: • S: Δdx= ? Δtx= ? Δty= ? vf = ? Δd = viΔt + ½ aΔt2 Δd = ½aΔt2 t = t = 4.0 s

  6. 2d a 2 (78.4 m) (-9.8m/s2) Δt = Example 2 • Horizontal Vertical A stone is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high. How far from the cliff does the stone hit the ground ? Δdy= 78.4 m vi = 0 m/s a = -9.8 m/s2 vx = 5.0 m/s • G: • U: • E: • S: • S: Δt = 4.0 s Δdx= ? Δtx= ? Δty= ? vf = ? Δd = vit + ½ a Δt2 Δd = ½ a Δt2 Δt = v = Δd/Δt Δd= vΔt Δd= (5.0 m/s)●(4.0 s) Δd= 20 m Δt = 4.0 s

  7. Example 3 A steel ball rolls with a constant velocity across a tabletop 0.950 m high. It rolls off the table and hits the ground 0.352 m from the edge of the table. How fast was the ball rolling just as it left the table? Δdx= 0.352 m vx = ? Δtx= ? vi = 0 m/s g = -9.8 m/s2 Δdy= 0.950 m Δty= ? vf = ?

  8. 2d a 2 (0.950m) (-9.8m/s2) Δ t = Example 3 (continued) • Horizontal Vertical A steel ball rolls with a constant velocity across a tabletop 0.950 m high. It rolls off the table and hits the ground 0.352 m from the edge of the table. How fast was the ball rolling just as it left the table? Δdy= 0.950 m vi = 0 m/s a = -9.8 m/s2 Δdx= 0.352 m • G: • U: • E: • S: • S: Δtx= ? vx = ? = 0.44 s Δty= ? vf = ? Δd = vi Δt + ½ a Δt2 Δ d = ½ a Δt2 Δ t = v = Δd/Δt vx= (0.352 m) / (0.44 s) vx = 0.8 m/s Δ t = 0.44 s

  9. Example 4 Divers at Acapulco dive horizontally from a cliff that is 65 meters high. If the rocks below the cliff protrude 27 meters beyond the edge of the cliff, what is the minimum horizontal velocity needed to safely clear the rocks below? Δdx= 27 m vx = ? Δtx= ? vi = 0 m/s g = -9.8 m/s2 Δdy= 65 m Δty= ? vf = ?

  10. Δ2d a 2 (65 m) (-9.8m/s2) Δt = Example 4 (continued) • Horizontal Vertical Divers at Acapulco dive horizontally from a cliff that is 65 meters high. If the rocks below the cliff protrude 27 meters beyond the edge of the cliff, what is the minimum horizontal velocity needed to safely clear the rocks below? Δdy= 65 m vi = 0 m/s a = -9.8 m/s2 Δdx= 27 m • G: • U: • E: • S: • S: Δtx= ? vx = ? = 3.64 s Δty= ? vf = ? Δ d = vi Δt + ½ a Δt2 Δ d = ½ at2 Δt = v = d/t vx = (27 m) / (3.64 s) vx = 7.43 m/s Δt = 3.64 s

  11. Example 5: • A tennis player stretches out to reach a ball that is just barely above the ground and successfully 'lobs' it over her opponent's head. The ball is hit with a speed of 18.7 m/s at an angle of 65.1 degrees. • a. Determine the time that the ball is in the air. • b. Determine the maximum height which the ball reaches. • c. Determine the distance the ball travels horizontally before landing.

  12. Example 5 – part a: • Horizontal Vertical V65.1o= 18.7 m/s Δdy= ? vi = ? a = -9.8 m/s2 • G: • U: • E: • S: • S: vx = ? Δtx= ? Δty= ? vf = ? vx = cos(65.1o)* 18.7 vy= sin(65.1o )* 18.7 vy= 16.96 m/s vf = vi + at t = vf– vi _______ a t = 0m/s– 16.96 m/s _____________________ -9.8m/s2 vx = 7.87 m/s ty = 1.73 s ttotal = 3.46 s

  13. Example 5 – part b: • Horizontal Vertical V65.1o= 18.7 m/s Δdy= ? vi = 16.96 m/s a = -9.8 m/s2 • G: • U: • E: • S: • S: Δtx= 3.46 s vx = 7.87 m/s Δty= 1.73 s vf = ? Δdx = ? d = v*t Δ d = vi Δt + ½ a Δt2 Δ d = ½ at2 Δ d = ½ (-9.8)(1.73)2 d = 7.87 m/s * 3.46 s d = 27.3 m Δ d = 14.7 m

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