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Physics: Post your doubts here!

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This is from a post that is two pages back, iin case you have any more doubts from this paper, you may find the answers there and on the page after that (p. 578 and p. 579):

Q14) When the projectile reaches it's highest point, it has a purely horizontal velocity and no vertical component of velocity.
Additionally, since there is no force on the projectile in the horizontal direction, it has the same horizontal velocity at it's highest point as it did when it was shot.

A little more on this - suppose the initial velocity has a magnitude of "v", the initial kinetic energy E = 1/2 * m * v² = 0.5mv².

Now, onto the focus of the question: since the initial velocity points at an angle 45° to the horizontal, the component of velocity in the horizontal direction should be

v(horizontal) = v * cos(45) = v/(√2)

This is, by our discussion above, the velocity of the projectile at it's maximum height. So, the kinetic energy at it's greatest height is

E(f) = (1/2) * m * (v/(√2))² = 0.5m * v²/2 = 0.25mv².

Let's divide E(f) by E, giving us

E(f)/E = (0.25mv²)/(0.5mv²) = 1/2
Therefore, E(f) = (1/2)E = 0.5E = A.

Good Luck for all your papers!
 
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Time wasnt sufficient... made some random guesses ... overall it was fine.. neither very good.. nor very bad :)
 
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Bro, I am an external student in my school.Today when I was solving the question paper, the Exam Officer and Supervisor repeatedly were disturbing me by sliding the OMR sheet between my vision and at times in between my pen's tip. This happened 5 to 6 times which caused loss of time. So what should I do to revert this act by the authorities.
 
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Heya guys! My paper was great...... but I won't get above 35 :(
Shits :(
I wanna screw myself................... TIME ZYADA Q NI DETE :( ????????????????
I knew answers but TIME!!! HELL WITH YOU . _ .
 
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Damn, just did Paper 12 and there were few questions repeated, I'm gonna honestly say this paper wasn't smooth with me :/ It wasn't easy and wasn't hard.
Anyways, good luck all!
 
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