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## Optimal angle for a projectile

# Optimal angle for a projectile part 2: Hangtime

## 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.