Lab 5

Projectile Motion

Eleanor Roosevelt High School

Mr. Chin-Sung Lin

Calculation of Projectile Motion

Example: A projectile was fired with initial velocity v0 horizontally from a cliff d meters above the ground. Calculate the horizontal range R of the projectile.

g

R

d

v0

t

Strategies of Solving Projectile Problems

H & V motions can be calculated independently H & V kinematics equations share the same

variable t

g

R

d

v0

t

Strategies of Solving Projectile Problems

H motion: dx = vx t R = v0 t

V motion: dy = d = 1/2 g t2 t = sqrt(2d/g)

So, R = v0 t = v0 * sqrt(2d/g)

g

R

d

v0

t

Numerical Example of Projectile Motion

H motion: dx = vx t R = v0 t = 10 t

V motion: dy = d = 1/2 g t2 t = sqrt(2 *19.62/9.81) = 2 s

So, R = v0 t = v0 * sqrt(2d/g) = 10 * 2 = 20 m

g = 9.81 m/s2

R

19.62 m

V0 = 10 m/s

t

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/s2

20 m

d

V0 = 10 m/s

t

Exercise 1: Projectile Problem

H motion: dx = vx t 20 = v0 t = 10 t t = 2 s

V motion: dy = d = 1/2 g t2 = 1/2 (9.81) 22 = 19.62 m

So, d = 19.62 m

g = 9.81 m/s2

20 m

d

V0 = 10 m/s

t

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 v0 of the projectile.

g = 9.81 m/s2

20 m

19.62 m

V0

t

Exercise 2: Projectile Problem

H motion: dx = vx t 20 = v0 t

V motion: dy = d = 1/2 g t2 t = sqrt(2 *19.62/9.81) = 2 s

So, 20 = v0 t = 2 v0 v0 = 20/2 = 10 m/s

g = 9.81 m/s2

20 m

19.62 m

V0

t