Download presentation

1
**Mathematical Application**

Projectile Motion Mathematical Application

2
**Simultaneous Equations**

Example: ๐ฃ ๐ฅ = โ ๐ ๐ฅ โ๐ก โ ๐ ๐ฆ = ๐ฃ ๐ โ๐ก+ 1 2 ๐โ ๐ก 2 If the two equations above are simultaneous equations, then their common variable (โ๐ก) 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: ๐=๐+๐ ๐=๐โ๐ Substitution: ๐=(๐)โ๐+๐ Solution: ๐=๐ Part II: ๐=๐+๐ ๐= ๐โ๐ +๐ (Substitution Method) ๐=๐โ๐+๐

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 ฮdx = ? vx = 5 m/s ฮ tx = ? vi = 0 m/s g = -9.8 m/s2 ฮdy = 78.4 m ฮty = ? ฮtP = ฮty = ฮtx vf = ?

5
**Example 1 (continued) Horizontal Vertical G: U: E: S:**

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? Horizontal Vertical ฮdy = 78.4 m vi = 0 m/s a = -9.8 m/s2 Vx = 5.0 m/s G: U: E: S: ฮdx = ? ฮtx = ? ฮty = ? vf = ? ฮd = viฮt + ยฝ aฮt2 ฮd = ยฝaฮt2 t = 2ฮd a t = 2 (78.4 m) (-9.8m/s2) t = 4.0 s

6
**Example 2 Horizontal Vertical G: U: E: S:**

A stone is thrown horizontally at a speed of 5.0 m/s from the top of a cliff m high. How far from the cliff does the stone hit the ground ? Horizontal Vertical ฮdy = 78.4 m vi = 0 m/s a = -9.8 m/s2 vx = 5.0 m/s G: U: E: S: ฮt = 4.0 s ฮdx = ? ฮtx = ? ฮty = ? vf = ? ฮd = vit + ยฝ a ฮt2 ฮd = ยฝ a ฮt2 ฮt = v = ฮd/ฮt ฮd = vฮt 2d a ฮd = (5.0 m/s)โ(4.0 s) ฮt = 2 (78.4 m) (-9.8m/s2) ฮd = 20 m ฮt = 4.0 s

7
**Example 3 ฮdx = 0.352 m vx = ? ฮtx = ? vi = 0 m/s g = -9.8 m/s2**

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? ฮdx = m vx = ? ฮtx = ? vi = 0 m/s g = -9.8 m/s2 ฮdy = m ฮty = ? vf = ?

8
**Example 3 (continued) Horizontal Vertical G: U: E: S: ฮ t = 0.44 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? Horizontal Vertical ฮdy = m vi = 0 m/s a = -9.8 m/s2 ฮdx = m G: U: E: S: ฮtx = ? vx = ? = 0.44 s ฮty = ? vf = ? ฮd = vi ฮt + ยฝ a ฮt2 ฮ d = ยฝ a ฮt2 ฮ t = v = ฮd/ฮt 2d a vx = (0.352 m) / (0.44 s) ฮ t = 2 (0.950m) (-9.8m/s2) vx = 0.8 m/s ฮ 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? ฮdx = 27 m vx = ? ฮtx = ? vi = 0 m/s g = -9.8 m/s2 ฮdy = 65 m ฮty = ? vf = ?

10
**Example 4 (continued) Horizontal Vertical G: U: E: 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? Horizontal Vertical ฮdy = 65 m vi = 0 m/s a = -9.8 m/s2 ฮdx = 27 m G: U: E: S: ฮtx = ? vx = ? = 3.64 s ฮty = ? vf = ? ฮ d = vi ฮt + ยฝ a ฮt2 ฮ d = ยฝ at2 ฮt = v = d/t ฮ2d a vx = (27 m) / (3.64 s) ฮt = 2 (65 m) (-9.8m/s2) vx = 7.43 m/s ฮt = 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 โ part a: Horizontal Vertical G: U: E: S: ty = 1.73 s**

V65.1o = 18.7 m/s ฮdy = ? vi = ? a = -9.8 m/s2 G: U: E: S: vx = ? ฮtx = ? ฮty = ? vf = ? vx = cos(65.1o) * 18.7 vy = sin(65.1o )* 18.7 vy = m/s vf = vi + at t = vf โ vi _______ a t = 0m/s โ m/s _____________________ -9.8m/s2 vx = 7.87 m/s ty = 1.73 s ttotal = 3.46 s

13
**Example 5 โ part b: Horizontal Vertical G: U: E: S: ฮ d = 14.7 m**

V65.1o = 18.7 m/s ฮdy = ? vi = m/s a = -9.8 m/s2 G: U: E: S: ฮtx = 3.46 s vx = 7.87 m/s ฮty = 1.73 s vf = ? ฮdx = ? d = v*t ฮ d = vi ฮt + ยฝ a ฮt2 ฮ d = ยฝ at2 ฮ d = ยฝ (-9.8)(1.73)2 d = 7.87 m/s * 3.46 s d = 27.3 m ฮ d = 14.7 m

Similar presentations

Presentation is loading. Please wait....

OK

2D Motion.

2D Motion.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on new technology in electrical engineering Ppt on wireless networking technology Ppt on boilers operations director Ppt on basic proportionality theorem for class 10 Ppt on magic maths Ppt on content development software Ppt on school education in india Ppt on computer languages memes Ppt on quality education definition Ppt on science and technology in india