Main content

## Physics library

### Course: Physics library > Unit 2

Lesson 2: Optimal angle for a projectile- Optimal angle for a projectile part 1: Components of initial velocity
- Optimal angle for a projectile part 2: Hangtime
- Optimal angle for a projectile part 3: Horizontal distance as a function of angle (and speed)
- Optimal angle for a projectile part 4: Finding the optimal angle and distance with a bit of calculus

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Optimal angle for a projectile part 2: Hangtime

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

## Want to join the conversation?

- 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 .(8 votes)
- 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!(17 votes)

- 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) ?(8 votes)
- 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**(14 votes)

- Why do we use Sv instead of S to find the hangtime at about2:30? Also how did the video get so clear all of sudden???(5 votes)
- "if thrown at the same velocity" - Brian.ruiz

Just to clarify everybody, he actually means "same vertical velocity"(7 votes)

- How come Sal is now using speed (scalar) as opposed to velocity (vector) as he has been for all of projectile motion?(7 votes)
- 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! :)(3 votes)

- Wait, so this is what's getting me confused:

I understand that the horizontal component is scos(theta) and the vertical component is ssin(theta), but why is time(a) Sv/g * 2?

The problem l have, is that the vertical component of the speed of the optimal angle is limited, and not at its peak. How could we calculate the time even though we're not at the maximum peak of the parabola? l'm afraid that l'm confusing people.(2 votes)- v = a * t

t = v / a

Also the units work as well. Dividing m/s by m/s^2 is the same as m/s * s^2/m (just the reciprocal of m/s^2). You would get s. The time in the air is only affected by the vertical component of the vector (s*sin(theta)), because gravity only acts on the vertical velocity. From the equations above, you can calculate t = v / a. In this case, Sv/g, but this is only calculating half of the parabola (the time it takes the object to go from the initial location to the location where it stops moving or when v = 0 because of the gravity). The total time in the air will be twice the time it takes to stop moving, so Sv/g*2 is the total time it stays in the air. Since gravity is constant, it accelerates the object downward at a constant rate.

I hope I am not confusing you.(6 votes)

- 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!(2 votes)
- The units are m/s^2. There are 9.8 of them You don't square the 9.8.(3 votes)

- how come there is no unit test for 2-D motion?(3 votes)
- 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?(3 votes)
- If you are given only the magnitude without the direction (the angle) you won't be able to calculate the horizontal distance.(1 vote)

- At2:00, 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?(1 vote)- 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(3 votes)

- 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?(2 votes)

## 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.