t = (2 * S * sin theta) / g (where g = -9.8 m/s^2) produces a negative number. from my understanding of 'Projectile At An Angle' video, this should be ( -2 * S * sin theta ) /g .

Max, an interesting point indeed! Here's the hidden thing! t=(2*S*sin<theta>) / g is actually derived from the equation; v = u + at ...Just put up v = 0 and u = S*sin<theta> And the tricky thing : a = -g (ie a= -9.8 m/s^2) .....solving the equation using these values gives us the time equation! The main thing is "a" is negative....g's value in the former equation should be positive because the sign was already taken into account while deriving the equation..so we just need the values... Hope it helped!

Could someone please enlighten me how did Sal derive the time in the air as the (2 times the initial velocity)/ (the acceleration due to gravity) ?

Well, imagine a ball thrown vertically upwards at a speed of 20 m/s. Let's assume that gravity exerts an acceleration of 10 m/s^2 downwards. Therefore, gravity would be initially working against the ball's velocity by slowing it down. After one second, the ball's speed would only be 10m/s, and it would take another second for the ball's speed to become zero. A total of two seconds has now elapsed. This can also be solved by *dividing the initial velocity by the acceleration due to gravity*. Afterwards, gravity will start to accelerate the ball downwards. It would also take another two seconds for the ball's speed to reach 20 m/s. By then, it would have already hit the ground. As you can see, the ball takes some time to get to its highest position (where speed is zero), then it also takes the same amount of time to fall back to the surface. Therefore, the formula for time is *2(initial velocity)/acceleration due to gravity*

How come Sal is now using speed (scalar) as opposed to velocity (vector) as he has been for all of projectile motion?

I think it's because Sal hasn't defined theta just as yet, therefore, the "vector" (used loosely in this context) only has magnitude but not direction. That's just my understanding. Some brilliant mind is bound to jump in and enlighten us all! :)

For the example at the end, if you were typing that in on a calculator, how do you account for m/s and m/s2 (squared). Would you do 9.8 squared or just 9.8? (9.8 being gravity at the bottom of the division). Thank you! such a great video!

The units are m/s^2. There are 9.8 of them You don't square the 9.8.

At 2:00, how do i know, that i takes exactly the same amount of time to fall down, as it takes to go up? Can someone tell me how to prove it mathematically?

To get the time it takes for the object to go up until its velocity is 0: _FinalVelocity = initialVelocity + (acceleration * time)_ which means 0 = 10 + (-10 * t) -10 = -10t t = 1 Now let's find out how far in the air it went: _Displacement = averageVelocity * time_ _averageVelocity = (finalVelocity + initialVelocity) / 2_ aV = (0 + 10) / 2 = 5 D = 5 * 1 D = 5 The object has to come the same distance down, so using that as our displacement, how long does it take to come down? _Displacement = (initialVelocity * time) + (.5 * acceleration * time^2)_ -5 = (0 * t) + (.5 * -10 * t^2) = -5 * t^2 t^2 = 1 t = 1

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Physics library

Course: Physics library > Unit 2

Lesson 2: Optimal angle for a projectile

Optimal angle for a projectile part 2: Hangtime

Name: Optimal angle for a projectile part 2: Hangtime
Uploaded: 2011-02-20T16:49:31Z
Description: Now it's time to think about hangtime for our far-flung projectile

Google Classroom

Now it's time to think about hangtime for our far-flung projectile. Created by Sal Khan.

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Maxfield
Posted 12 years ago. Direct link to Maxfield's post “t = (2 * S * sin theta) /...”
t = (2 * S * sin theta) / g (where g = -9.8 m/s^2) produces a negative number. from my understanding of 'Projectile At An Angle' video, this should be ( -2 * S * sin theta ) /g .
Button navigates to signup pageComment on Maxfield's post “t = (2 * S * sin theta) /...”
(8 votes)
Answer
- Himalaya Agarwal
  Posted 11 years ago. Direct link to Himalaya Agarwal's post “Max, an interesting point...”
  Max, an interesting point indeed!
  Here's the hidden thing!
  t=(2*S*sin<theta>) / g is actually derived from the equation;
  v = u + at ...Just put up v = 0 and u = S*sin<theta>
  And the tricky thing : a = -g (ie a= -9.8 m/s^2) .....solving the equation using these values gives us the time equation!
  The main thing is "a" is negative....g's value in the former equation should be positive because the sign was already taken into account while deriving the equation..so we just need the values...
  Hope it helped!
  Button navigates to signup page
  (17 votes)
HemaPrasath
Posted 11 years ago. Direct link to HemaPrasath's post “Could someone please enli...”
Could someone please enlighten me how did Sal derive the time in the air as the (2 times the initial velocity)/ (the acceleration due to gravity) ?
Button navigates to signup pageButton navigates to signup page
(8 votes)
Answer
- Paolo Miguel Bartolo
  Posted 11 years ago. Direct link to Paolo Miguel Bartolo's post “Well, imagine a ball thro...”
  Well, imagine a ball thrown vertically upwards at a speed of 20 m/s. Let's assume that gravity exerts an acceleration of 10 m/s^2 downwards. Therefore, gravity would be initially working against the ball's velocity by slowing it down. After one second, the ball's speed would only be 10m/s, and it would take another second for the ball's speed to become zero. A total of two seconds has now elapsed. This can also be solved by dividing the initial velocity by the acceleration due to gravity.
  
  Afterwards, gravity will start to accelerate the ball downwards. It would also take another two seconds for the ball's speed to reach 20 m/s. By then, it would have already hit the ground.
  
  As you can see, the ball takes some time to get to its highest position (where speed is zero), then it also takes the same amount of time to fall back to the surface. Therefore, the formula for time is 2(initial velocity)/acceleration due to gravity
  Comment on Paolo Miguel Bartolo's post “Well, imagine a ball thro...”
  (14 votes)
Albert Hanan
Posted 9 years ago. Direct link to Albert Hanan's post “How come Sal is now using...”
How come Sal is now using speed (scalar) as opposed to velocity (vector) as he has been for all of projectile motion?
Button navigates to signup pageButton navigates to signup page
(7 votes)
Answer
- S Chung
  Posted 8 years ago. Direct link to S Chung's post “I think it's because Sal ...”
  I think it's because Sal hasn't defined theta just as yet, therefore, the "vector" (used loosely in this context) only has magnitude but not direction.
  
  That's just my understanding. Some brilliant mind is bound to jump in and enlighten us all! :)
  Button navigates to signup page
  (3 votes)
izzyh99
Posted 9 years ago. Direct link to izzyh99's post “For the example at the en...”
For the example at the end, if you were typing that in on a calculator, how do you account for m/s and m/s2 (squared). Would you do 9.8 squared or just 9.8? (9.8 being gravity at the bottom of the division). Thank you! such a great video!
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Andrew M
  Posted 9 years ago. Direct link to Andrew M's post “The units are m/s^2. The...”
  The units are m/s^2. There are 9.8 of them You don't square the 9.8.
  Button navigates to signup page
  (3 votes)
rajesh ranjan
Posted 4 years ago. Direct link to rajesh ranjan's post “how come there is no unit...”
how come there is no unit test for 2-D motion?
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(3 votes)
Answer
Michael Michell
Posted 10 years ago. Direct link to Michael Michell's post “If you are given the velo...”
If you are given the velocity vector and the problem asks you to find how far the projectile is being thrown. How do you calculate the horizontal distance if you are not given an angle?
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(3 votes)
Answer
- Prince Grey
  Posted 5 years ago. Direct link to Prince Grey's post “If you are given only the...”
  If you are given only the magnitude without the direction (the angle) you won't be able to calculate the horizontal distance.
  Button navigates to signup page
  (1 vote)
Imran Hussain
Posted 8 years ago. Direct link to Imran Hussain's post “i dont understand why you...”
i dont understand why you added the times 2 to the equation
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(2 votes)
Answer
- NPav
  Posted 7 years ago. Direct link to NPav's post “Because the first part of...”
  Because the first part of the equation only accounts for the upward direction, and because what goes up must also come back down, we multiply by two to see the time when it lands back on the ground.
  Button navigates to signup page
  (1 vote)
Martin L
Posted 9 years ago. Direct link to Martin L's post “At 2:00, how do i know, t...”
At
2:00
, how do i know, that i takes exactly the same amount of time to fall down, as it takes to go up?
Can someone tell me how to prove it mathematically?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Mr. Kloud7
  Posted 9 years ago. Direct link to Mr. Kloud7's post “To get the time it takes ...”
  To get the time it takes for the object to go up until its velocity is 0:
  FinalVelocity = initialVelocity + (acceleration * time) which means
  0 = 10 + (-10 * t)
  -10 = -10t
  t = 1
  Now let's find out how far in the air it went:
  Displacement = averageVelocity * time
  averageVelocity = (finalVelocity + initialVelocity) / 2
  aV = (0 + 10) / 2 = 5
  D = 5 * 1
  D = 5
  The object has to come the same distance down, so using that as our displacement, how long does it take to come down?
  Displacement = (initialVelocity * time) + (.5 * acceleration * time^2)
  -5 = (0 * t) + (.5 * -10 * t^2) = -5 * t^2
  t^2 = 1
  t = 1
  Button navigates to signup page
  (3 votes)
Vincent Simon
Posted 8 years ago. Direct link to Vincent Simon's post “So when using this formul...”
So when using this formula: ta = Sv/g *2 wouldn't the resulting time always be negative because the gravitational force is negative?
Shouldn't it rather be ta = -Sv/g *2 to get reasonable results?
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(2 votes)
Answer
Mohadeseh Ahmadi
Posted 10 years ago. Direct link to Mohadeseh Ahmadi's post “What is the difference be...”
What is the difference between this problem and the one he solved in the ‘projectile at an angle video’ problem?
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(2 votes)
Answer
- Zech.Tan911
  Posted 10 years ago. Direct link to Zech.Tan911's post “This problem is more like...”
  This problem is more like an extension of the projectile at an angle problem. The tools are the same--you use sine and cosine to find the velocity vector components. He's just demonstrating that it is possible to find the angle that gives you the furthest distance using these equations.
  Comment on Zech.Tan911's post “This problem is more like...”
  (1 vote)

Video transcript

Let's figure out how long this object is going to be in the air given that its vertical velocity, or the magnitude of the vertical velocity is s sine of theta. So its speed in the vertical direction is s sine of theta. So how long is it going to be in the air? Well if I told you that something is going upwards at 10 meters per second and gravity is decelerating it at 10 meters per second squared. So every second it's going to slow it down by 10 meters per second, how long will it take for that object to get to 0, to stop moving? Let me write that down. Let's say that some object is moving upwards at 10 meters per second. And let's say that the gravity is slowing it down. Slowing it down at 10 meters per second. So every second that goes by, it'll slow this thing down by 10 meters per second. Well, it'll take it exactly 1 second to make it go from 10 meters per second to 0 meters per second. And then it's going to be at some height in the air, and then the thing's going to start accelerating. Gravity is going to start accelerating it downward. And then it'll take another second for it to go from 0-- from having no velocity, to having 10 meters per second again. So in this case, it'll take the time in the air-- we could say time sub air, I guess, is going to be equal to this 10 meters per second, your velocity. 10 meters per second divided by the acceleration. Divided by this 10 meters per second. 10 meters per second times 2. This is how long it'll take for the object to go from 10 meters per second to 0 at some point in the air. And then it's going to take the exact same amount of time for it to fall back to the ground. So times 2. If the object was moving upwards at 20 meters per second and gravity is still slowing it down at 10 meters per second per second, then it's going to take 2 seconds. If this was 20, then this would be 20. And it'll take 2 seconds to slow it down to 0 and then 2 more seconds until it hits the ground again. For it to speed back up as it approaches the ground. So no matter what your upward velocity, the time in the air is going to be your speed, your vertical speed, divided by the acceleration of gravity. And this is the amount of time it's going to take you to go from this point to that point. To have some vertical velocity and then slow down to 0. And it's going to take the exact same amount of time for you to speed back up by gravity and get to your original speed. We're assuming no air resistance. So it's kind of a pure problem. So this is the time up, the time down is going to be the same thing. So we can multiply that by 2. Now, we already know what the vertical component for our problem is. It is s sine of theta. So we could just substitute that back in there. And we know how long we're going to be in the air. The time in the air is going to be our speed-- or I should maybe put the 2 out front. 2 times s sine theta. Let me make it clear. This 2 right here is this 2 right there. All of that over the acceleration of gravity. So if you told me that I'm shooting this object off at-- I don't know-- 100 meters per second. So if this is 100 meters per second and if theta were-- I don't know-- let's say theta were 30 degrees, then sine of theta would be 1/2. So it'd be 100 meters per second times 1/2 divided by the acceleration of gravity times 2 would tell you exactly how long you would be in the air. How long it takes to go up all the way, become stationery, and then fall back down to the ground.