Presentation is loading. Please wait.

Presentation is loading. Please wait.

Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration.

Similar presentations


Presentation on theme: "Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration."— Presentation transcript:

1 Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration.

2 Free Fall Really, just a special case of one dimensional kinematics. Uses the same equations. – But, since displacement is vertical, it is written as y (instead of x). – Acceleration is no longer a variable, it is a fixed value 'g'.

3 Free Fall Free Fall: The only acceleration or force is due to a constant gravity. All objects moving under the influence of “only gravity” are said to be in free fall. Free fall acceleration does not depend on the object’s original motion, or on the mass. Acceleration is the one thing constant to all falling objects. It is constant for the entire time they are in the air.

4 Free Fall Free Fall: The only acceleration or force is due to a constant gravity. This means that the significant figures of force due to other forces like Air Resistance are InSignificant.

5 Freely Falling Objects A freely falling object is any object moving under the influence of gravity alone. Acceleration does not depend upon the initial motion of the object!!!!!!! Examples of Freefall with different initial motions. – Dropped –released from rest – Thrown downward – Thrown upward – Throwing a ball Up or DOWN AFTER the ball leaves the hands. Examples of NonFreefall or NOT FREEFALL. – Throwing a ball Up or DOWN BEFORE the ball leaves the hands, while the hands are still pushing on it is NOT freefall. – A rocket engine is still burning and pushing up on a rocket.

6 Freely Falling Objects A freely falling object is any object moving under the influence of gravity alone. Acceleration does not depend upon the initial motion of the object!!!!!!! Examples of Freefall with different initial motions. – Dropped –released from rest – Thrown downward – Thrown upward – Throwing a ball Up or DOWN AFTER the ball leaves the hands. So what does initial motion affect then??? It changes the v i in v f = v i + at. It does not change the a.

7 Acceleration of Freely Falling Object The acceleration of an object in free fall is directed downward, regardless of the initial motion The magnitude of free fall acceleration is g= 9.80 m/s 2 which is the number used in your textbooks, so use 9.80 so that you match the black and white textbooks. Only valid on the Earth’s surface at OUR ALTITUDE. Not to be confused with g for grams British textbooks use 9.81 m/s 2 instead of 9.80 m/s 2 like American textbooks. Why?

8 Acceleration of Freely Falling Object The acceleration of an object in free fall is directed downward, regardless of the initial motion The magnitude of free fall acceleration is g= 9.80 m/s 2 which is the number used in your textbooks, so use 9.80 so that you match the black and white textbooks. Only valid on the Earth’s surface at OUR ALTITUDE. Not to be confused with g for grams GRAVITY VARIES, because of what?

9 Strength of gravity The value of 'g' is affected by altitude. At the top of Mt. Everest, g = m/s 2 (.028 % less)

10 Acceleration of Free Fall, cont. We will neglect air resistance Free fall motion is constantly accelerated motion in ONLY one dimension, which is y. DO NOT USE a x = m/s 2 Let upward be positive Use the kinematic equations with a y = -g= m/s 2

11 Acceleration of Free Fall, cont. Use the kinematic equations with a= g= m/s 2 REMEMBER: + means up, - means down Write these equations on page: Kinematics Equation Page (Choose 1 side or other.) Y Axis y = y i + v i t - ½ gt 2 v f = v i - g t v f 2 = v i 2 -2g Δy g= 9.80 m/s 2 (Because a = g) Y Axis y = y i + v i t + ½ at 2 v f = v i + a t v f 2 = v i 2 + 2a Δy g= m/s 2 (Because g = Neg)

12 Free Fall –an object dropped If I drop a soccer ball off the roof of a very, very tall building, what is the velocity 8 seconds later? What is the displacement? Solve these in your powerpoint notes.

13 Free Fall –an object dropped V f = v i - g t V 8 = 0 - (9.80 m/s 2 )(8 s) V 8 = m/s What is the displacement? y 8 = y i + v i t - ½ gt 2 y 8 = (8) - ½ (9.80 m/s 2 )(8 s) 2 y 8 = -313 m What’s wrong with these answers? v f = v i + a t v 8 = 0 + (-9.80 m/s 2 )(8 s) v 8 = m/s What is the displacement? y 8 = y i + v i t + ½ at 2 y 8 = (8)+½ (-9.80 m/s 2 )(8 s) 2 y 8 = -313 m If I drop a soccer ball off the roof of a very, very tall building, what is the velocity 8 seconds later?

14 Free Fall –an object dropped V f = v i - g t V 8 = 0 - (9.80 m/s 2 )(8 s) V 8 = m/s What is the displacement? y 8 = y i + v i t - ½ gt 2 y 8 = (8) - ½ (9.80 m/s 2 )(8 s) 2 y 8 = -313 m What’s wrong with these answers? Sig Fig. Ans: -80 and -300 v f = v i + a t v 8 = 0 + (-9.80 m/s 2 )(8 s) v 8 = m/s What is the displacement? y 8 = y i + v i t + ½ at 2 y 8 = (8)+½ (-9.80 m/s 2 )(8 s) 2 y 8 = -313 m If I drop a soccer ball off the roof of a very, very tall building, what is the velocity 8 seconds later?

15 Free Fall of an object thrown upward Initial velocity is upward, so positive The instantaneous velocity at the maximum height is zero. V y=max = 0 a y = -g = m/s 2 everywhere in the motion: at the start, top, and end And everywhere inbetween. v i ≠0 a y = -g At Top v y = 0 a y = -g

16 Free Fall -object thrown upward V i > 0 a y = -g At Top v y = 0 a y = -g A soccer ball is kicked directly upwards with a velocity of 8 m/s. At what time will it’s velocity be zero? At what time will its height peak? What’s the highest it will go? How much time will it spend in the air before it hits the ground? With what velocity will it strike the ground?

17 Free Fall -object thrown upward V i > 0 a y = -g At Top v y = 0 a y = -g A soccer ball is kicked directly upwards with a velocity of 8 m/s. At what time will it’s velocity be zero? Ans: v f = v i + a t  t=.816s =.8s At what time will its height peak? Ans: Same Question. Notice v y = 0 At the top. What’s the highest it will go? Ans: y f = y i + v i t + ½ at 2  3.27m = 3m Or use v f 2 = v i 2 + 2a Δy How much time will it spend in the air before it hits the ground? Double 1 st ans t = 1.6s With what velocity will it strike the ground? Discuss ball, bullets, arrows, and air resistance.

18 thrown upwards, free fall only same height The motion may be symmetrical IF it lands at the same height as thrown. Then t up = t down Then v f = -v i Break the motion into two parts up and down

19 Free Fall –an object thrown downward a y = -g= m/s 2 Initial velocity ≠0 Since up is positive, initial velocity will be negative. V o < 0 a = -g

20 Free Fall --object thrown Downward Same Equations, but V o will be negative A person fires a handgun straight downwards from a helicopter. The muzzle velocity of the gun is 300 m/s. The helicopter is 250 m above the ground. With what velocity will the bullet hit the ground?

21 Free Fall --object thrown Downward Same Equations, but V o will be negative A person fires a handgun straight downwards from a helicopter. The muzzle velocity of the gun is 300 m/s. The helicopter is 250 m above the ground. With what velocity will the bullet hit the ground? ANS: v f 2 = v i 2 + 2a Δy = -308 m/s All v i,a, Δy = negative v f = choose neg sq.root.

22 Problem Solving tips Think about and understand the situation GIVENS: List the given information. REMEMBER g = m/s 2 and at the top of the arc, v y = 0 (y direction only.) DIAGRAM: Make a quick drawing of the situation FIND: List the unknown you are solving for. EQUATIONS: List the equations used to solve. CALCULATIONS: Show your work. Think about units SOLUTION: List your answer. Focus on the expected result Think about what a reasonable answer should be

23 Example We throw a bowling ball upwards from the edge of the roof. What happens at points B, C and E?

24 Example Initial velocity at A is upward (+) and acceleration is -g(-9.80 m/s 2 ) At B, the velocity is 0 and the acceleration is -g(-9.80 m/s 2 ) At C, the velocity has the same magnitude as at A, but is in the opposite direction The displacement is –50.0 m (it ends up 50.0 m below its starting point)

25 Problem Solving – Question: If a flea can jump.440 m high, what is the initial velocity as it leaves the ground?

26 Problem Solving – Question: If a flea can jump.440 m, what is the initial velocity as it leaves the ground? y = y i +v i t + ½ at 2 v = v i + a t v f 2 = v i 2 +2ay

27 Problem Solving – Question: If a flea can jump.440 m, what is the initial velocity as it leaves the ground? Just look at flea on the way up. y is.440 m v i is what we’re looking for a is -9.8 m/s 2 v f is zero, since we’re looking at the top of his jump.

28 Problem Solving – Question: If a flea can jump.440 m, what is the initial velocity as it leaves the ground? Just look at flea on the way up. y is.440 m v i is what we’re looking for a is -9.8 m/s 2 v f is zero, since we’re looking at the top of his jump. v f 2 = v i 2 +2ay

29 Problem Solving – Question: If a flea can jump.440 m, what is the initial velocity as it leaves the ground? 0 = v i (-9.80m/s 2 )(.440m) v f 2 = v i 2 +2ay

30 Problem Solving – Question: If a flea can jump.440 m, what is the initial velocity as it leaves the ground? v f 2 = v i a y 0 = v i (-9.80m/s 2 )(.440m) = v i m 2 /s 2 so v i = + / m/s Which makes sense, + or -?

31 Problem Solving – Question: If a flea can jump.440 m, how much time does he spend going up? y f = y i +v i t + ½ at 2 v f = v i + a t v f 2 = v i 2 +2ay

32 Problem Solving – Question: If a flea can jump.440 m, how much time does he spend going up? a is -9.8 m/s 2 v i = 2.94 m/s v f = m/s 2 = ( m/s) / t t =.300 sec v f = v i + a t

33 Problem Solving – What is the velocity when he lands? Just look at the problem on the way down. v f 2 = 0 + (-9.80m/s 2 )(-.440m) = m 2 /s 2 v f = +/ m/s Which makes sense, + or - ? v f 2 = v i 2 +ay

34 Problem Solving How much time does he spend in the air? y f = y i +v i t + ½ at 2 v f = v i + a t v f 2 = v i 2 +2ay

35 Problem Solving How much time does he spend in the air? -2.94m/s = 2.94m/s + (-9.81m/s2)*t -5.82m/s = -9.81m/s2*t t =.600 sec Twice the time it took to go up v f = v i + a t


Download ppt "Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration."

Similar presentations


Ads by Google