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Mechanics projectile motion

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For e) why is the direction 50 degrees "below the horizontal" ?

It looks like it's above
Original post by G.Y


For e) why is the direction 50 degrees "below the horizontal" ?

It looks like it's above


If you mark on tbe velocity at the very end, you would see that it is pointing downward hence it is going below the horizontal and the angle they’re referring to is the opposite of the one you’re seeing.
Original post by G.Y


For e) why is the direction 50 degrees "below the horizontal" ?

It looks like it's above


Because the motion of the ball follows a quadratic curve and is therefore symmetrical about a vertical line drawn through the point where the ball is at its maximum height.
Just as the genius said, it is because in this case we assume that the curve is symmetrical in its journey. So, the first half of the journey of the projectile matches the second half and so does the curve. So, half way through its journey at its maximum point when v = 0, the curve reflects itself. So, at the start, before leaving the ground, the ball was projected at 50 degrees so at the end, before it reaches the ground, it will be travelling at 50 degrees too. The velocity will also be the same too.

Original post by old_engineer
Because the motion of the ball follows a quadratic curve and is therefore symmetrical about a vertical line drawn through the point where the ball is at its maximum height.
Also, the reason for this is because the acceleration is constant. If the acceleration is constant, the velocity is constant assuming no air resistance. Therefore, the journey will be symmetrical. That's why the time, distance, everything is symmetrical - it's all to do with the acceleration.
Reply 5
Original post by Chittesh14
Also, the reason for this is because the acceleration is constant. If the acceleration is constant, the velocity is constant assuming no air resistance. Therefore, the journey will be symmetrical. That's why the time, distance, everything is symmetrical - it's all to do with the acceleration.


By velocity is constant surely you mean symmetrical for both sides? It does vary over the course of the motion?
Original post by G.Y
By velocity is constant surely you mean symmetrical for both sides? It does vary over the course of the motion?


Sorry I mean the velocity decreases at a constant rate.

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