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## Physics library

### Course: Physics library>Unit 1

Lesson 5: Old videos on projectile motion

# Projectile motion (part 2)

A derivation of a new motion equation. Created by Sal Khan.

## Want to join the conversation?

• At the change in the distance you said is 500m but when we throw it up a bit by a velocity of 30m/s the change in distance will change as it goes up. It has to be 500+(the distance it went up) isn't it this way? i checked the equation you solved and the answer is correct but i am not getting the point i just mentioned,
Thank you. •   Too be more clear Sal should mention that we are not really here talking about distance(a scalar quantity) we are actually talking about the "displacement"(a vector) of the penny i.e. How far away the penny is when it lands on the ground compared to where it was at the instant we initiallt threw it. Thats why it is correct to use 500m and ignore any additional added height caused by throwing the penny upwards(this height is already accounted for in the equation by the Vi term having the opposite sign to Vf.

No matter how high you thow the penny up, in the end its displacement from the point at which it started will still me 500m or rather -500m (since up was defined as positive and down as negative).
• How far does the penny go up in the air before gravity takes over and it begins it's decent? 45m? • Indeed. That's exactly what you get with the given assumptions (like a = g = -10m/s^2), and make use of a simple fact, which is that at the highest point, right at that instant the velocity is going to be zero. So Vf = 0, and you solve for d, and there you go... 45m.
• By solving in the end, we got 104 m/s. Yet, the penny is moving downwards so it should be -104 m/s. But, in the actual math, the negatives cancelled out and we were left with a positive integer. So to indicate downward motion, do we just put a negative in front of the 104, or should we have gotten a -104 at the end of the math? The math Sal did wasn't wrong, so I am confused. • While solving for the final velocity, it is obvious that the penny is going downwards (conventionally represented as negative). Sal found out that Vf^2=10900. He took its square root to get approximately 104 m/s. The square root of a positive number can be both positive and negative. For example, 2^2=4 and (-2)^2=4. In this case Sal should have taken the negative square root of 10900, not the principal root. So, the final velocity is definitely -104 m/s or 104 m/s downwards.
Hope this helped.
• At Sal calculates the Velocity final = 104m/s (down)
If a gun fired a penny straight up from the bottom of the cliff at 104m/s (up) would it achieve the same height as the penny Sal threw from the top of the cliff? • Should the velocity be -104 m/s? • The final velocity is 104 m/s. Working out the equations, I was wondering why the answer is not a negative number, since final velocity is going down by convention. • Like you said at the end of the video, the object goes up first, and then it comes down. Therefore, how can delta d be 500m/s? That's just the height of the cliff, but you didn't take into account the distance that the ball travels upwards. How would you do that? • Maybe it would help you if you tried using the equation for throwing the penny downwards, with velocity -30m/s.
Surprise! The result will be still 104m/s. So in fact throwing the penny up at 30m/s will result in the penny coming back to the starting position (0m) moving at the velocity -30m/s. Because we are not using t in this equation, we can "ignore" the upwards movement.
At least that's how i understand it, correct me if I'm wrong.
• doesn't weight matter when you fall?
(1 vote) • Sir,
Does the weight of the coin matter according to its mass?  