Tossing a projectile over a wall beside a platform

In summary, given a platform with a height of 4m and a brick wall with height Y on the edge of it, a launch site some distance behind the wall, and a landing site past the wall and beyond the platform, with a launch angle of 40 degrees, the range can be found using the formula x=vot cos @, where @ = 40 degrees. At the highest point, when the velocity is 0, the displacement in the X-direction is 83.39m. The height of Y at this point is 15.46m, and adding 4m to this gives a total height of 19.46m. The displacement in the X-direction from the maximum point to the landing site
  • #1
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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...
 
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  • #2
Could you provide the exact wording of the question please?
 
  • #3
physicdiagramcopy.jpg


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

Displacement in the X-direction:

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

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

Y=Vot-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=Vot= (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?
 
  • #4
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.
 
  • #5
I have absolutely no idea, I'm afraid. As far as I'm concerned, it's 105.02m away. :p
 

1. How do I calculate the trajectory of a projectile when tossing it over a wall beside a platform?

The trajectory of a projectile can be calculated using the following formula: y = y0 + xtan(θ) - (gx2)/(2v02cos2(θ)), where y is the height of the projectile at any given point, y0 is the initial height of the projectile, x is the horizontal distance traveled, θ is the angle at which the projectile is launched, g is the acceleration due to gravity (9.8 m/s2), and v0 is the initial velocity of the projectile.

2. How does the height of the wall and platform affect the trajectory of the projectile?

The height of the wall and platform will affect the trajectory of the projectile by changing the initial height y0 in the calculation. A higher wall or platform will result in a higher initial height and therefore a higher trajectory. The angle at which the projectile is launched may also need to be adjusted to account for the height of the wall and platform.

3. What is the optimal angle to launch the projectile at to clear the wall and land on the platform?

The optimal angle to launch the projectile at will depend on the height of the wall and platform, as well as the initial velocity of the projectile. To determine the optimal angle, the following equation can be used: θ = tan-1((2gy0)/(v02)). This equation will give the angle at which the projectile will travel the furthest distance and clear the wall while landing on the platform.

4. How does the initial velocity of the projectile impact its trajectory when tossing it over a wall beside a platform?

The initial velocity of the projectile will greatly impact its trajectory. A higher initial velocity will result in a higher trajectory and a longer distance traveled. The initial velocity will also affect the optimal angle at which the projectile should be launched to clear the wall and land on the platform.

5. Is there a way to increase the distance the projectile travels when tossing it over a wall beside a platform?

There are a few ways to increase the distance the projectile travels when tossing it over a wall beside a platform. One option is to increase the initial velocity of the projectile. Another option is to adjust the angle at which the projectile is launched to find the optimal angle for maximum distance. Additionally, the height of the wall and platform can be adjusted to allow for a higher initial height and therefore a longer trajectory.

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