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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

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 .
• 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.
• how come there is no unit test for 2-D motion?
• 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?
• If you are given only the magnitude without the direction (the angle) you won't be able to calculate the horizontal distance.
(1 vote)
• i dont understand why you added the times 2 to the equation
• 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.
(1 vote)
• At , 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