Question on Translational Motion

[phatty925]

I have a question on translation motion:

I understand that that the horizontal distance travelled (range) is dictated by the horizontal velocity. so to find range, you would multiply the horizontal velocity by the time the projectile is in the air.
However, I have difficulty finding "time in air."
I understand that time in air is dictated by the vertical velocity. this would mean that time in air=vertical distance/vertical velocity.
what value do you use for vertical distance to find the time in air?
Can someone please explain this using an example with specific numbers? Thanks!

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[yalla22]

Originally posted by phatty925
I have a question on translation motion:

I understand that that the horizontal distance travelled (range) is dictated by the horizontal velocity. so to find range, you would multiply the horizontal velocity by the time the projectile is in the air.
However, I have difficulty finding "time in air."
I understand that time in air is dictated by the vertical velocity. this would mean that time in air=vertical distance/vertical velocity.
what value do you use for vertical distance to find the time in air?
Can someone please explain this using an example with specific numbers? Thanks!

hey phatty,
"time in air=vertical distance/vertical velocity" is not correct.see in the vertical dimension velocity is changing due to the force of gravity-at the top of the projectiles motion Vy is 0 while at other times it obv. isnt. gravity changes the v and so you can not use that equation. that one is only for the horizontal dimension where gravity is not considered. its difficult to give an example as numbers vary per problem but let me try...
so in these problems with a x and y dimension you are many times given the Vi and an angle of take off. using sin/cos laws you can find the vy and vx values, right? so then using vfy=viy+at one can find half of the "air time". viy is what you calculated and vfy is o-multiply by 2 to get total air time (since that is the time to get to top of projectile motion but what you want is time to reach delta y=0)
if you give a specific prob, i can help you.
yalla

 

[heifetz]

Time in air


Imagine a ball thrown up into the air. its velocity starts from some value say 20m/s and when it reaches the top of its flight its vertical velocity is zero. then it starts to fall and goes from the top at 0m/s and gradually accelerates until it hits the ground and it hit the ground at -20m/s assuming no air resistance. so you see the ball starts and ends with the same speed.

So i like to divide my motion into 2 parts. 1st part is from the start to the top and the 2nd part from the top to the bottom. Find the time for the first part and then mutiple it by 2 because the 2nd half of the motion mirrors the first exactly. So assuming you know your trans. motion equation you can use
initial velocity (20m/s)
final velocity (0m/s)
accel -10 m/s

then figure out the time for your 1st half of the motion and mutliply that time by 2 to get the total time in the air.

hope that helps.

 

[Energon]

phatty understand your concern.

You are right on the 1st part of range being dependent upon Vx. This is because Vx is contstant throughout the projectile motion. However, this is not the case for vertical motion. It goes from whatever the starting y component is to 0 at the top and back to what it started with on its landing.

To find total time spent in air I like to use the formula:

t = 2 (Vsin@/g)

It makes it much simpler.

eg: A ball is thrown into the air at a speed of 40 m/s and an angle of 30 degrees. How much time does the ball spend in the air before it hits the ground?

V Sin30 = 40(1/2)= 20m/s

t(total) = 2 [(20m/s)/ g] = 4s

However this only applies for a projectile motion problem; not for a ball that is thrown straight up in the air like the one heifetz mentioned, that would be linear motion.

 

[phatty925]

thanks, guys! that really helps.
but i have a couple of more questions: AMMD, when you said that your explanation couldn't be applied to a situation like the one given by heifetz where you throw a ball straight up in the air, you just mean that there is no sin theta to consider, right? but you can still use t=2(v/g), right?
for example, if a ball is thrown up into the air with a velocity of 20m/s, the total time the ball is in the air before hitting the ground would be 2(20/10), which equals 4 seconds. is this correct?

also, when do you use x=1/2at^2 then? do you use this after you half the time in air to find how high the object goes? could you guys give me an example of when to use this equation?

thanks!

 

[heifetz]

x = 1/2at^2 is simply the short hand form of x = Vot + 1/2 at^2. you use this short version when your intial velocity is zero leaving you with x = 1/2at^2. Most likely you would use this equation when an object is dropped from a building or plane with a zero starting velocity and solve for displacement.

example) a man on top of a tall building drops a baseball and in 1 second it falls 1 story from the top. in one more second it will be?
a)2 stories below the top
b)3 stories from the top
c) 4 stories from the top
d)16 stories from the top

use the above formula to solve this.

hope that helps

 

[Energon]

Phatty,

For the eg of the ball thrown up in the air:
in order to calculate the time taken for the entire motion, yes you can use the formula:

t = 2 v/g.
However, since most of these questions also entail the height (h)of the ball thrown or dropped; I like to use:

t = 2 (2h/g)^1/2

eg. Jerry who was standing on the roof of his house 20 m high (I dont know if he was peeping in to the neighbors bedroom j/k), dropped a ball to Ben standing on the lawn. How long did it take for the ball to fall and what was its final velocity?
Solution:
t(1/2) = (2h/g)^1/2
t (1/2) = (40/10)^1/2 = 2s.
for Vf, I like to use:
Vf = (2gh)^1/2
Vf = (400)^1/2 = 20m/s

Memorizing these hybrid formulas just saves time and enables me to nail the easy questions.
hope this helped

AMMD

 

[Energon]

Heifetz, for your problem:
example) a man on top of a tall building drops a baseball and in 1 second it falls 1 story from the top. in one more second it will be?
a)2 stories below the top
b)3 stories from the top
c) 4 stories from the top
d)16 stories from the top

I figure out using t= (2h/g)^1/2 that the ball fell 20m in 2s. Using the same equation I also see that one floor = 5m. THus in the 2nd second the ball is at the 4th story from the top.
Is that correct?

 

[Mudd]

Originally posted by AMMD
I figure out using t= (2h/g)^1/2 that the ball fell 20m in 2s. Using the same equation I also see that one floor = 5m. THus in the 2nd second the ball is at the 4th story from the top.
Is that correct?

That is correct. For that question, you can also use the "sqaures rules". d is proportional to t^2 and there is no initial velocity in the y-direction, so d at any time t can be found by the ratio of t^2 values.

If it were to drop three floors in 2 seconds, then it drops twelve floors total in 4 seconds, b/c 4^2/2^2 is 4, so the total d is four times greater, which is 4 x 3 =12.

It's based on the idea that if you do the same type of question using the same equations enough, you'll see the same pattern and thus can skip steps.