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Given: a platform which is 4m high. On the edge of it is a brick wall with height Y. The launch site is some way behind the wall. The landing site is past the wall and beyond the platform. The launch angle is 40 degrees. Find the range.
x=vot cos @ (where @=angle)
y=vot sin @
range=(vo^2 sin 2@)/g
Take up as positive.
First, find x at highest point when v=0.
Vy=vo-gt
0=30-8.31.t
30=8.31.t
t=3.61
What is delta x at time 3.61?
delta x=vot cos 40
delta x=(30)(3.61)(0.77)
delta x=(108.3)(0.77)
delta x=83.39 at the highest point, at the top of the brick wall and at the end of the platform.
Now, find height of Y at that point:
y=vot sin 40-(gt^2)/2
y=(30)(3.61)(0.643)-[(8.31)(g.31)^2]/2
y=69.6-54.14
y=15.46m
Now, add 4m to this:
15.46+4=19.46m
Find length of delta x from maximum to landing site when y=19.46 at maximum:
and, er, this is where things begin to trail off...
x=vot cos @ (where @=angle)
y=vot sin @
range=(vo^2 sin 2@)/g
Take up as positive.
First, find x at highest point when v=0.
Vy=vo-gt
0=30-8.31.t
30=8.31.t
t=3.61
What is delta x at time 3.61?
delta x=vot cos 40
delta x=(30)(3.61)(0.77)
delta x=(108.3)(0.77)
delta x=83.39 at the highest point, at the top of the brick wall and at the end of the platform.
Now, find height of Y at that point:
y=vot sin 40-(gt^2)/2
y=(30)(3.61)(0.643)-[(8.31)(g.31)^2]/2
y=69.6-54.14
y=15.46m
Now, add 4m to this:
15.46+4=19.46m
Find length of delta x from maximum to landing site when y=19.46 at maximum:
and, er, this is where things begin to trail off...