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1. Today in physics I was working on the last question. I have an answer but it does not fit the requirements of the question. It goes as follows:
A particle is projected from a point 0 on horizontal ground, with a speed of 35m/s and an angle of projection of '@' (Theta). It strikes the ground again at a point 75m from 0. Find the two possible values of @.

So to start I broke down the vector of 35 into [email protected] horizontally and [email protected] vertically. I then used suvat on the horizontal plane to get time as 15/([email protected]).Knowing at half this time 15/([email protected]) the projectile would be at its peak vertically I then used suvat on the vertical plane to find @.

Ill try to make it as clear as possible. Starting with the formula V=u+at
0=([email protected]) -9.81(15/[[email protected]])
([email protected])([email protected])/15=9.81
490([email protected]@)/15=245([email protected])/15=9.81You can then proceed to rearrange the last statement for @.Which gives a value of around 18.4.Is this correct? where is the second value?
2. You can go both directions,
18.4 degrees would go 75m (lets say) left you can also go 75m to the right.
3. (Original post by Squirtle.U)
You can go both directions,
18.4 degrees would go 75m (lets say) left you can also go 75m to the right.
Seeing as 75m is the displacement, we know the direction and so there would only be one direction in which its fired?
4. (Original post by billdjango99)
Seeing as 75m is the displacement, we know the direction and so there would only be one direction in which its fired?
Is it described as a displacement in the question or a point from O.

If as you stated:
"It strikes the ground again at a point 75m from 0 "
that makes it a distance and direction is not implied.
5. (Original post by Squirtle.U)
Is it described as a displacement in the question or a point from O.

If as you stated:
"It strikes the ground again at a point 75m from 0 "
that makes it a distance and direction is not implied.
of course ahah, thank you!
6. (Original post by billdjango99)
of course ahah, thank you!
not so fast...

This might not have been pointed out but the farthest you can get with a projectile under A level assumptions (no air drag etc.) is when you project it at 45 degrees - and that's the one unique solution for that speed and distance, therefore any angle other than 45 degrees has two solutions that'll reach the same distance. both in the same direction - or of course 4 if you allow the cheaty solutions in the opposite direction.

Spoiler:
Show
easy answer is 0+θ and 90-θ

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Updated: October 10, 2016
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