Warm - up Problem:

1. Warm - up Problem: A sprinter accelerates from rest to 10.0 m/s in 1.35s. a.) What is her acceleration? b.) How far has she traveled in 1.35 seconds? a = 7.41 m/s2 d = 6.75 m

2. Freefall Motion • You will be able to successfully complete the following: • Develop a relationship between gravitational effect on displacement in 1-D vertical motion. • Evaluate velocity at different time intervals throughout freefall. • Compare the 1-D Horizontal motion equations (with constant acceleration) with those of 1-D Vertical motion. • Calculate unknown variables using above equations. • Identify critical points along the path of the freefalling object.

3. Such an object has an acceleration on Earth of -9.81 m/s/s, downward. g = -9.81 m/s2 This acceleration due to gravity, “g” can be substituted into any of the kinematic equations for “a” A free-falling object is an object which is falling under the sole influence of gravity. Thus, any object which is moving and being acted upon only by the force of gravity is said to be "in a state of free fall."

4. t = 0.00 s t = 1.00 s t = 2.00 s t = 3.00 s t = 4.00 s Since the effects of gravity affect objects the same in the upward direction as the downward direction, let’s first examine an object falling in the downward direction. Remember direction is very important with vectors. Especially for calculations!!!  A 100.0 gram ball is dropped from a cliff and takes 4.00 seconds to hit the ground. Evaluate the velocity at each 1.00 second interval. Draw a diagram to support your answer. vf = 0.00 m/s vf = vi + g t vf = -9.81 m/s vf = -19.62 m/s vf = -29.43 m/s vf = -39.24 m/s

5. Since we are discussing vertical motion we will use y as the symbol for vertical displacement. “y” can be substituted into kinematic equations for “x”

6. From the previous problem, calculate the distance traveled at each of the time intervals. Use g = -9.81 m/s2 0 0 0 0 0 0 t = 0.00 s t = 1.00 s t = 2.00 s t = 3.00 s t = 4.00 s Δ y = vi t + ½ g t2 = 0.00 m = - 4.91 m Δ y = vi t + ½ (-9.81 m/s2) (1.00)2 = - 19.6 m Δ y = vi t + ½ (- 9.81 m/s2) (2.00)2 = - 44.2 m Δ y = vi t + ½ (- 9.81 m/s2) (3.00)2 = - 78.5 m Δ y = vi t + ½ (- 9.81 m/s2) (4.00)2

7. Example: Falling from a tower. Suppose that a bowling ball is dropped from a tower 70.0 m high. How far will it have fallen after 1.00 s, 2.00 s, and 3.00 s? (Neglect air resistance. Use g = -9.81 m/s2) t = 0.00s t = 1.00s h = 70.0 m t = 2.00s t = 3.00s = 0.00 m Δ y = vi t + ½ (-9.81 m/s2) (0.00)2 = - 4.91 m Δ y = vi t + ½ (-9.81 m/s2) (1.00)2 = - 19.6 m Δ y = vi t + ½ (-9.81 m/s2) (2.00)2 = - 44.2 m Δ y = vi t + ½ (-9.81 m/s2) (3.00)2

8. Now, without completely frying your brain cells, let’s go to freefall in the upward direction. **Remember, Upward is (+) and Downward is (-). Gravity is gravity, and it will behave the same way as it did in the downward descent. g is still – 9.81 m/s2

9. = 2.0 seconds Problem: A coin is thrown upward with an initial velocity of 19.6 m/s. If g = -9.81 m/s2, how long will the coin take to reach its peak? Draw a diagram to support your answer. What is the final velocity at the peak? vtop = 0.00 m/s So, using the equation, vf = vi + g t 0.00 m/s = 19.6 m/s + (- 9.81 m/s2) ( t ) t = 0.00 s vi = 19.6 m/s

10. t = 0.00 s vi = 19.6 m/s Using the same problem, calculate the height of the coin once it reaches its peak. (Hint: Δ y = vi t + ½ g t2 ) Height = Δ y = vi t + ½ g t2 Height = (19.6 m/s)(2.00 s) + ½ (-9.81 m/s2)(2.00 s)2 So, height = 19.6 m

11. Problem: A bullet is shot straight up with an exiting velocity of 370. m/s. Use g = -9.81 m/s2. Neglect air resistance. A.) How long does it take to reach its maximum height? B.) How high will the bullet go before making its way back down to the earth? vf = vi + g t t = 37.7 s h ~ 6,980 m Height = Δ y = vi t + ½ g t2