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Projectile Motion Mathematical Application. Simultaneous Equations.

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Presentation on theme: "Projectile Motion Mathematical Application. Simultaneous Equations."— Presentation transcript:

1 Projectile Motion Mathematical Application

2 Simultaneous Equations

3 Substitution Method 1.choose one of the variables common for both equations 2.manipulate one of the equations to solve for that variable 3.combine the two equations by substituting the manipulated equality for the same variable in the second equation. 4.Solve the combined equation for the second unknown variable

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 v x = 5 m/s Δd y = 78.4 m v i = 0 m/s g = -9.8 m/s 2 v f = ? Δt y = ? Δd x = ? Δ t x = ? Δt y = Δt x Δt P =

5 Example 1 (continued) 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? G: U: E: S: Horizontal Vertical V x = 5.0 m/s Δd x = ? Δt x = ? Δd y = 78.4 m v i = 0 m/s a = -9.8 m/s 2 Δt y = ? v f = ? Δd = v i Δt + ½ aΔt 2 Δd = ½aΔt 2 t = 2Δd a t = 2 (78.4 m) (-9.8m/s 2 ) t = 4.0 s

6 Example 2 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 ? G: U: E: S: Horizontal Vertical v x = 5.0 m/s Δd x = ? Δt x = ? Δd y = 78.4 m v i = 0 m/s a = -9.8 m/s 2 Δt y = ? v f = ? Δd = v i t + ½ a Δt 2 Δd = ½ a Δt 2 Δt = 2d a Δt = 2 (78.4 m) (-9.8m/s 2 ) Δt = 4.0 s v = Δd/Δt Δd = vΔt Δd = (5.0 m/s)●(4.0 s) Δd = 20 m

7 Example 3 A steel ball rolls with a constant velocity across a tabletop m high. It rolls off the table and hits the ground m from the edge of the table. How fast was the ball rolling just as it left the table? v x = ? Δd y = m v i = 0 m/s g = -9.8 m/s 2 v f = ? Δt y = ? Δd x = m Δt x = ?

8 Example 3 (continued) G: U: E: S: Horizontal Vertical Δd x = m Δt x = ? v x = ? Δd y = m v i = 0 m/s a = -9.8 m/s 2 Δt y = ? v f = ? Δd = v i Δt + ½ a Δt 2 Δ d = ½ a Δt 2 Δ t = 2d a Δ t = 2 (0.950m) (-9.8m/s 2 ) Δ t = 0.44 s v = Δd/Δt v x = (0.352 m) / (0.44 s) v x = 0.8 m/s A steel ball rolls with a constant velocity across a tabletop m high. It rolls off the table and hits the ground m from the edge of the table. How fast was the ball rolling just as it left the table? = 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? v x = ? Δd y = 65 m v i = 0 m/s g = -9.8 m/s 2 v f = ? Δt y = ? Δd x = 27 m Δt x = ?

10 Example 4 (continued) G: U: E: S: Horizontal Vertical Δd x = 27 m Δt x = ? v x = ? Δd y = 65 m v i = 0 m/s a = -9.8 m/s 2 Δt y = ? v f = ? Δ d = v i Δt + ½ a Δt 2 Δ d = ½ at 2 Δt = Δ2d a Δt = 2 (65 m) (-9.8m/s 2 ) Δt = 3.64 s v = d/t v x = (27 m) / (3.64 s) v x = 7.43 m/s = 3.64 s 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?

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: G: U: E: S: Horizontal Vertical v x = ? Δt x = ? Δd y = ? v i = ? a = -9.8 m/s 2 Δt y = ? v f = ? v y = sin(65.1 o )* 18.7 v y = m/s t y = 1.73 s v x = cos(65.1 o ) * 18.7 v x = 7.87 m/s V 65.1 o = 18.7 m/s v f = v i + at t = v f – v i _______ a t = 0m/s – m/s _____________________ -9.8m/s 2 t total = 3.46 s

13 Example 5 – part b: G: U: E: S: Horizontal Vertical Δt x = 3.46 s v x = 7.87 m/s Δd y = ? v i = m/s a = -9.8 m/s 2 Δt y = 1.73 s v f = ? Δd x = ? d = 27.3 m V 65.1 o = 18.7 m/s Δ d = v i Δt + ½ a Δt 2 Δ d = ½ at 2 Δ d = ½ (-9.8)(1.73) 2 Δ d = 14.7 m d = v*t d = 7.87 m/s * 3.46 s


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