Tossing a projectile over a wall beside a platform

[paperclip]

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=volt cos @ (where @=angle)
y=volt 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=volt 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=volt 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...

 

[cristo]

Could you provide the exact wording of the question please?

 

[paperclip]

Here's another stab at it:
Y=30 sin 40=19.28m
X=30 cos 40=22.98

Displacement in the X-direction:

x=volt+ 0.5 at^2, a=o, so cancel out (0.5 at)
x=volt
x= (22.98) (t)

Need to find time when Y=-4, taking up as positive:

Y=volt-0.5gt^2
-4=30(t)-0.5(8.314)(t^2)(multiply both sides by half)
-8=30t-8.314t^2
8+30t=8.314t^2 (t cancels out)
8+30=8.314t
38/8.314=t, t = 4.57

FIND X WHEN t=-4:
x=volt= (22.98)(4.57)
x=105.02m

Therefore the launch site needs to be 105.02m away from the target in order to hit it.

Does this seem correct to you?

 

[denverdoc]

I still don't understand the problem. Where is the target? Anywhere on the other side of the wall or are you tring to maximize distance from the wall? Minimize it?

If the wall is some arbitrary height,Y, I would think the final answer would not be purely numerical answer but expressed in terms of Y.

 

[paperclip]

I have absolutely no idea, I'm afraid. As far as I'm concerned, it's 105.02m away. :p