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M2 Projectile Motion watch

1. Thingy is projected from edge of a cliff at 20ms-1 @ 30 degrees elevation.
After T Seconds , Angle of depression is 45 degrees.

Find T.
2. What is the vertical component of the velocity when the angle of depression is 45 degrees?

hint:
Spoiler:
Show
Spoiler:
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The horizontal velocity is constant, you know theta and so you can work out the vertical component.

When you've got that, consider motion vertically to find time
3. (Original post by JAKstriked)
What is the vertical component of the velocity when the angle of depression is 45 degrees?

hint:
Spoiler:
Show
Spoiler:
Show

The horizontal velocity is constant, you know theta and so you can work out the vertical component.

When you've got that, consider motion vertically to find time
Ok. Since the angle is 45 degrees i am assuming that at time T, Vertical velocity = Horizontal Velocity.

Horizontal Velocity is 20cos30 = 10sqrt3
So at T Vertical velocity also should be 10sqrt3

So I apply my formula v = u + at
10sqrt3 = -10sin30 + 9.8t

Which gives me the wrong answer 2.78
What am I doing wrong here?
4. (Original post by antonfigo)
Ok. Since the angle is 45 degrees i am assuming that at time T, Vertical velocity = Horizontal Velocity.

Horizontal Velocity is 20cos30 = 10sqrt3
So at T Vertical velocity also should be 10sqrt3

So I apply my formula v = u + at
10sqrt3 = -10sin30 + 9.8t

Which gives me the wrong answer 2.78
What am I doing wrong here?
Well the initial velocity isn't -10sin30, it's -20sin30, but that's still not giving me the right answer
5. (Original post by antonfigo)
Thingy is projected from edge of a cliff at 20ms-1 @ 30 degrees elevation.
After T Seconds , Angle of depression is 45 degrees.

Find T.

Angle of depression is 45 degrees, means that the POSITION of "thingy" will be 45 degrees down from the origin, not that the velocity will be at an angle of 45 degrees down.
6. (Original post by ghostwalker)
Angle of depression is 45 degrees, means that the POSITION of "thingy" will be 45 degrees down from the origin, not that the velocity will be at an angle of 45 degrees down.
I got the answer. You were right...
I assumed that since 45 degrees, the horizontal and vertical displacement is equal.

Applying s = ut + .5at^2 for both hori and verti

Hence 20cos30t = -20sin30t + 4.9t
So t is 5.59 s

Thanks PPL

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