1. Lab 5 Projectile Motion Eleanor Roosevelt High School Mr. Chin-Sung Lin

2. Calculation of Projectile Motion  Example: A projectile was fired with initial velocity v 0 horizontally from a cliff d meters above the ground. Calculate the horizontal range R of the projectile. g R d v0v0 t

3. Strategies of Solving Projectile Problems  H & V motions can be calculated independently  H & V kinematics equations share the same variable t g R d v0v0 t

4. Strategies of Solving Projectile Problems H motion: d x = v x t R = v 0 t V motion: d y = d = 1 / 2 g t 2 t = sqrt(2d/g) So, R = v 0 t = v 0 * sqrt(2d/g) g R d v0v0 t

5. Numerical Example of Projectile Motion H motion: d x = v x t R = v 0 t = 10 t V motion: d y = d = 1 / 2 g t 2 t = sqrt(2 *19.62/9.81) = 2 s So, R = v 0 t = v 0 * sqrt(2d/g) = 10 * 2 = 20 m g = 9.81 m/s 2 R 19.62 m V 0 = 10 m/s t

6. Exercise 1: Projectile Problem A projectile was fired with initial velocity 10 m/s horizontally from a cliff. If the horizontal range of the projectile is 20 m, calculate the height d of the cliff. g = 9.81 m/s 2 20 m d V 0 = 10 m/s t

7. Exercise 1: Projectile Problem H motion: d x = v x t 20 = v 0 t = 10 t t = 2 s V motion: d y = d = 1 / 2 g t 2 = 1 / 2 (9.81) 2 2 = 19.62 m So, d = 19.62 m g = 9.81 m/s 2 20 m d V 0 = 10 m/s t

8. Exercise 2: Projectile Problem A projectile was fired horizontally from a cliff 19.62 m above the ground. If the horizontal range of the projectile is 20 m, calculate the initial velocity v 0 of the projectile. g = 9.81 m/s 2 20 m 19.62 m V0V0 t

9. Exercise 2: Projectile Problem H motion: d x = v x t 20 = v 0 t V motion: d y = d = 1 / 2 g t 2 t = sqrt(2 *19.62/9.81) = 2 s So, 20 = v 0 t = 2 v 0 v 0 = 20/2 = 10 m/s g = 9.81 m/s 2 20 m 19.62 m V0V0 t