1. GIG
Read the passage and mark your answers on your whiteboard. NOT ON THE PAPER.
Questions

2. Projectile Motion at an Angle

3. Projectile Review Position and velocity formulae for projectiles
xf = xi + vixt
yf = yi +viyt - .5gt2
vxf = vxi (In other words, horizontal velocity does not change)
vyf = vyi - gt

4. Try this on your whiteboard
A ball rolls off a table with a horizontal velocity of 2.47m/s and hits the ground .350s later. How fast was it moving horizontally when it hit the ground? Vertically?

5. Try this on your whiteboard
A ball rolls off a table with a horizontal velocity of 2.47m/s and hits the ground .350s later. How fast was it moving horizontally when it hit the ground? Vertically?
Vfx = vix = 2.47m/s
Vfy = 0m/s m/s2(.350s) = -3.43m/s

6. Angled Projectiles
So far, we have only considered projectiles launched horizontally off a ledge. These projectiles have an initial vertical velocity of zero.
What if the projectile is launched with an initial vertical velocity?
This is where trigonometry comes back into play.

7. Angled Projectiles
Problems with an angled projectile typically tell you an object was launched with an initial velocity of ___ m/s and at an angle of ___ degrees. You will then calculate distance, time, or final velocity.

8. Position-Time Formulae with Angle θ
x-component
xf = xi + (vi cosθ)t
y-component
yf = yi + (vi sinθ)t - .5gt2

9. Throwing a Ball
An athlete throws a ball at 4.99 m/s at an angle 35.0° above the horizontal. After 1.00s it hits the ground. How high was the ball above the ground when the athlete released it? Assume gravity = 9.81m/s2.

10. Throwing a Ball
An athlete throws a ball at 4.99 m/s at an angle 35.0° above the horizontal. After 1.00s it hits the ground. How high was the ball above the ground when the athlete released it? Assume gravity = 9.81m/s2.
yf = yi + (vi sinθ)t - .5gt2
0m = yi + (4.99m/s)(sin35.0°)(1.00s) - .5(9.81m/s2)(1.00s)2
0m = yi m m yi = 2.04m

11. Try this on your whiteboard
A golf ball is launched at 13.5 m/s at an angle 54.0° above the horizontal. After 2.24s it lands. Assuming the height of the launch and landing spots is equal and gravity = 9.81m/s2, how far did the ball travel?

12. Try this on your whiteboard
A golf ball is launched at 13.5 m/s at an angle 54.0° above the horizontal. After 2.24s it lands. Assuming the height of the launch and landing spots is equal, how far did the ball travel?
xf = xi + (vi cosθ)t
xf = 0m + (13.5m/s)(cos54.0°)(2.24s) = 17.8m

13. Try this on your whiteboard - Part II
A golf ball is launched at 13.5 m/s at an angle 54.0° above the horizontal. How high above the ground was the ball at 1.76s? Assume gravity = 9.81m/s2.

14. Try this on your whiteboard - Part II
A golf ball is launched at 13.5 m/s at an angle 54.0° above the horizontal. How high above the ground was the ball at 1.76s? Assume gravity = 9.81m/s2.
yf = yi + (vi sinθ)t - .5gt2
yf = 0m + (13.5m/s)(sin54.0°)(1.76s) - .5(9.81m/s2)(1.76s)2
yf = 0m m m = 4.03m

15. Velocity-Time Formulae with Angle θ
x-component
vxf = vicosθ
y-component
Vyf = visinθ - gt

16. Jumping Frog
A frog leaps into the air at 2.5 m/s at an angle 52° above the horizontal. What is his vertical speed after 0.25s? What is the horizontal component of his speed?

17. Jumping Frog
A frog leaps into the air at 2.5 m/s at an angle 52° above the horizontal. What is his vertical speed after 0.25s? What is the horizontal component of his speed?
xvf = vicosθ = 2.5m/s(cos52°) = 1.5m/s
Note that time is irrelevant to his horizontal velocity. The horizontal velocity of a projectile is constant.

18. Jumping Frog
A frog leaps into the air at 2.5 m/s at an angle 52° above the horizontal. What is his vertical speed after 0.25s?

19. Jumping Frog
A frog leaps into the air at 2.5 m/s at an angle 52° above the horizontal. What is his vertical speed after 0.25s?
yvf = visinθ - gt = 2.5m/s(sin52°) m/s2(.25s)
yvf = -0.48m/s
Note that the negative sign means the frog is heading back down at this time.

20. Try this on your whiteboard - Part III
A golf ball is launched at 13.5 m/s at an angle 54.0° above the horizontal. What is its speed when time = 1.76s and vertical position = 4.03m? Assume gravity = 9.81m/s2.

21. Try this on your whiteboard - Part III
A golf ball is launched at 13.5 m/s at an angle 54.0° above the horizontal. What is its speed when time = 1.76s and vertical position = 4.03m? Assume gravity = 9.81m/s2.
You will need to solve the vertical and horizontal components of speed and then use the pythagorean theorem. The horizontal is not hard:
vxf = vicosθ vxf = 13.5m/s(cos54.0°) = m/s

22. Try this on your whiteboard - Part III
A golf ball is launched at 13.5 m/s at an angle 54.0° above the horizontal. What is its speed when time = 1.76s and vertical position = 4.03m? Assume gravity = 9.81m/s2.
Vertical component:
Vyf = visinθ - gt = 13.5m/s(sin54.0°) m/s2(1.76s)
Vyf = m/s m/s = m/s

23. Try this on your whiteboard - Part III
A golf ball is launched at 13.5 m/s at an angle 54.0° above the horizontal. What is its speed when time = 1.76s and vertical position = 4.03m? Assume gravity = 9.81m/s2.
Use pythagorean theorem to put the two components together.
(7.9351m/s)2 + ( m/s)2 = m2/s2
√ = 10.2m/s

24. Homework
2-D Motion Problems, due Friday.

25. Closure Answer on your whiteboard:
An archer shoots an arrow at a target 15.0 m away. Although aimed at the center of the target, the arrow lands .52m below it. How long was the arrow in flight?