Homework Statement
I was wondering, when an object's motion is in the shape of a parabola, and it lands at the same height that it was shot from.. why would the initial velocity and the angle from the ground at which it was shot from.. be equal to the final velocity at which it hit the ground...
Yes sorry i didn't clarify that..
Since both the vertical and horizontal have times that are the same, then I used acceleration formula (v=initial+at) manipulated, inserted what I knew and I got the time. Then, since this time is the same as the horizontal time, then I used it and the distance...
No problem, I appreciate your efforts and time trying to help me :)
So the horizontal, vertical velocity calculated using the trigonometry method in the previous post and in the method that I quoted above, both methods should yield the same answer right?
Using the angle mentioned in the question and the initial velocity which is the hypotenuse, I would do this and setup a right angled triangle?
and of course (Sine, Cosine, Tangent)
ok.. so the time that the object was in the air is the length of the whole event. Therefore, I would only use that time only to determine the velocity of the horizontal correct? (v=d/t) After this velocity is calculated, then I would use it and half of the horizontal distance to calculate the...
Thanks for your replies they have helped a lot.
I was just wondering... would the time at maximum height be half of the total time or not necessarily? I know that the distance at maximum height would be half of the horizontal distance.
Also, for the vertical the initial velocity would be...
Hello,
I am not completely sure how to attempt this problem but I have given it a shot. If anyone knows how to do this please look it over, or post your attempt. Thank you!
Homework Statement
An object was shot at an angle of 49 degrees and landed 60m away at the same height. The initial...