# Projectiles Problem- HELP!!

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#1
Attachment 670358Attachment 670358670360Attachment 670358670360Hey guys, sorry I am new to step and am finding myself confused even with the marking in front of me... D:

I am really stumped over this projectile question in the step handbook by Steven Skilos, is my diagram correct? can anyone please explain how i would draw the second diagram and what is going on? whats with all the ls and the vt in the final part for tan(theta) ? I think I'm approaching it wrong... Ive been struggling for a few days nowName:
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3 years ago
#2
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#3
(Original post by marinacalder)
Attachment 670358Attachment 670358670360Attachment 670358670360Hey guys, sorry I am new to step and am finding myself confused even with the marking in front of me... D:

I am really stumped over this projectile question in the step handbook by Steven Skilos, is my diagram correct? can anyone please explain how i would draw the second diagram and what is going on? whats with all the ls and the vt in the final part for tan(theta) ? I think I'm approaching it wrong... Ive been struggling for a few days nowName:
Attachment 670362670364
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#4
sorry
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3 years ago
#5
You seem a bit confused about theta and the triangle required. As the ball falls down on its parabolic path, draw a tangent at the relevant point. Make a triangle.

Theta lies in the bottom right corner, tangent = hypoteneuse, opposite = y and adjacent = d - x (the length between point x and point d forms the adjacent of theta's triangle).

For the second bit, point d isn't stationary, it's moving left towards point x. Distance moved is vt. Their difference forms your adjacent side: (d - vt) - x.

Where me and the markscheme differ is they have a +L. Regardless, derivative of tan(theta) is constant.
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#6
(Original post by Physics Enemy)
You seem a bit confused about theta and the triangle required. As the ball falls down on its parabolic path, draw a tangent at the relevant point. Make a triangle.

Theta lies in the bottom right corner, tangent = hypoteneuse, opposite = y and adjacent = d - x (the length between point x and point d forms the adjacent of theta's triangle).

For the second bit, point d isn't stationary, it's moving left towards point x. Distance moved is vt. Their difference forms your adjacent side: (d - vt) - x.

Where me and the markscheme differ is they have a +L. Regardless, derivative of tan(theta) is constant.
Ok, so in doing these questions what am I actually doing? Am I working with a co- ordinate system in terms of different variables for the same point? (sorry I'm self taught in mechanics)
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3 years ago
#7
(Original post by marinacalder)
Ok, so in doing these questions what am I actually doing? Am I working with a co- ordinate system in terms of different variables for the same point? (sorry I'm self taught in mechanics)
Just sketch parabolas on the same x-y grid from (0, 0). I was pointing out theta on the way up is bottom left /_ but on the way down bottom right _\

In both parts the hyp of the triangle for @ is tangent to the curve and in the velocity's direction.

For the last part, ball's landing point moves left over time. So range is (d - vt) instead of d. So [(d - vt) - x] is the adjacent of our triangle for @.

But the markscheme has a +L, unsure why. I haven't looked much into the Q though.
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