If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Physics library

### Course: Physics library>Unit 2

Lesson 2: Optimal angle for a projectile

# 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.

## 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.