help me with projectiles before i throw my laptop out of a window
A particle is projected such that its range over level ground is three times the maximum height of its path. Find the angle of projection.
^^ i'm not being funny i've tried to do this question for twenty minutes now and don't understand at all, someone please help
^^ i'm not being funny i've tried to do this question for twenty minutes now and don't understand at all, someone please help
Original post by emmab0488
A particle is projected such that its range over level ground is three times the maximum height of its path. Find the angle of projection.
^^ i'm not being funny i've tried to do this question for twenty minutes now and don't understand at all, someone please help
^^ i'm not being funny i've tried to do this question for twenty minutes now and don't understand at all, someone please help
Clearly you're looking to create an equation of the form "range = 3 x max height"
So, using the suvat equations, can you work out the values of range and max height in terms of the intial velocity and angle of projection? What have you been able to do so far?
(edited 5 months ago)
Original post by emmab0488
A particle is projected such that its range over level ground is three times the maximum height of its path. Find the angle of projection.
^^ i'm not being funny i've tried to do this question for twenty minutes now and don't understand at all, someone please help
^^ i'm not being funny i've tried to do this question for twenty minutes now and don't understand at all, someone please help
Something a bit more basic (geometry) than using the suvat equations (which is the correct way to go as in ghostwalkers post), you could note that the x-y trajectory is a parabola (where 2h is the max height) so
y = ax^2+bx+c
and fit the points (0,0), (6h,0),(3h,2h), where the last is the maximum so a completed square is an obvious way to go. Then b (gradient of the tangent at zero) would give tan of the initial angle of projection.
Or note that it drops 2h in the second half of the parabola, so it must drop the same height in first half from tne initial projection which gives the relevant (famous) triangle which corresponds to the inital gradient.
(edited 5 months ago)
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