Throwing a projectile so that it clears the window

In summary: Without knowing the angle, I cannot determine the initial velocity-self-contradictory, if you ask me.
  • #1
negation
818
0

Homework Statement



Standing on the ground 3.0 m from the wall of a building, you want to throw a package from your 1.6 m shoulder level to someone in a second-floor window 4.1 m above the ground

a)At what speed should you throw the package so it just barely reaches the window?
b) At what angle should you throw the package so it just barely reaches the window?

I know how to solve it conceptually but the workings are causing me problems mainly because I am asked to determine the 2 crucial unknowns.

The Attempt at a Solution



Assuming the projectile crosses the window at the apex of the trajectory.

vyf = vyi -gt
0 = vyi -9.8t
t = vi sin Θ/g

y(vi sin Θ/g) = y(0) +vi.t - 0.5gt^2
y(vi sin Θ/g) - y(0) = (vi sin Θ)(vi sin Θ/g) - 4.9(vi sin Θ/g)^2
2.7 = [(vi ^2 sin^2Θ)/g] -4.9[(vi^2 sin^Θ)/g^2]

I cannot assume Θ= 45 because part b asked for the angle. However, without knowing the angle, I cannot determine the initial velocity-self-contradictory, if you ask me.
 
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  • #2
It's not a very obvious question is it. I think it's looking for when it has no vertical speed, what else would barely reaches the window mean? In that case it's not to bad.
Vertical component:
s=2.5 u=? v=0 a=-9.81 t=? s=vt-0.5at^2, 2.5=-0.5*-9.81*t^2, t=0.714, also can find using other suvat equation u=7.

Solve for horizontal using t=0.714 find that u=4.20

overall magnitude=(7^2+4.2^2)^0.5=8.16.
Angle=tan^-1(7/4.2)=59 degrees.
 
  • #3
You don't need to assume an angle. You just need to break it up into 2 parts initial velocity in y, and then in x.

Initial velocity in y would the speed needed to reach 2.5m(from the shoulder). And x is a little more tricky, you need to realize that it will reach its apex half way through its journey in x, which you would want it to travel 6m before returning to the height it was thrown from.
 
  • #4
Darren93 said:
It's not a very obvious question is it. I think it's looking for when it has no vertical speed, what else would barely reaches the window mean? In that case it's not to bad.
Vertical component:
s=2.5 u=? v=0 a=-9.81 t=? s=vt-0.5at^2, 2.5=-0.5*-9.81*t^2, t=0.714, also can find using other suvat equation u=7.

Solve for horizontal using t=0.714 find that u=4.20

overall magnitude=(7^2+4.2^2)^0.5=8.16.
Angle=tan^-1(7/4.2)=59 degrees.


V=0? Is v the initial speed? If so, it doesn't make sense.
I'll work on it tonight.
 
  • #5
V= 0 would be the velocity at its apex.
 
  • #6
The velocity in y that is.
 
  • #7
MostlyHarmless said:
The velocity in y that is.


At the apex yes. That would be the final velocity using vyf^2 -vyi^2 =2g(yf-yi)
But the other poster was using the equation
Yf= yi + vyi.t -0.5gt^2 and denoting vyi =0
 
  • #8
Well, obviously you know that the initial velocity can't be 0. But yes, I would definitely use the formula you just posted, there is no reason to use time if you don't have to.
 
  • #9
MostlyHarmless said:
Well, obviously you know that the initial velocity can't be 0. But yes, I would definitely use the formula you just posted, there is no reason to use time if you don't have to.

0 = vyi^2 = 2(-9.8)(4.2-1.5)
vyi = 7.27 ms^-1

The answer to part (a) is apparently 8.3ms^-1

What is wrong here?
 
  • #10
That's only the y component, now find the x component.
 
  • #11
Also the numbers you used 4.2 and 1.6 should be 4.1 and 1.6 according to the op.
 
  • #12
MostlyHarmless said:
That's only the y component, now find the x component.

That explains. I've to find the y and x component and find the magnitude, right?
I think there's another way to determine the y-component and that is to find the height of the trajectory.
However, from sources, it says, in using the height trajectory formula, the time, t = vyi/g must be substitute into y = yaverage velocity x t. Doesn't it make more sense to sub time into y(t) = y(0) + vi.t - 0.5gt^2?
 
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  • #13
In the same time period, t = vi sinΘ/g, the projectile moves a horizontal distance of
x = vi.t = vi cosΘ(vi sinΘ/g), am I right?
 
  • #14
MostlyHarmless said:
Also the numbers you used 4.2 and 1.6 should be 4.1 and 1.6 according to the op.

In the same time period, t = vi sinΘ/g, the projectile moves a horizontal distance of
x = vi.t = vi cosΘ(vi sinΘ/g), am I right?
It should be right but apparently, there are too many unknown variables.

x = vi^2 sin(2Θ)/2g
x = 3m and g = 9.8 but I have no knowledge of Θ and vi

3 m = vi^2sin(2Θ)/2g
vi^2 sin(2Θ) = 58.8
Unable to proceed further.
 
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  • #15
It should be right but apparently, there are too many unknown variables.

You haven't used the height. I think if you write another equation using the height it will work out but I haven't tried.
 
  • #16
Edit
 
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  • #17
CWatters said:
You haven't used the height. I think if you write another equation using the height it will work out but I haven't tried.

It's the same thing. They both give me the same y-component value.
The tricky part is to solve for the x-component.
I know vyi = 7.27.
In order to determine the numerical time value, I sub vyi into y(t) = vi.t -0.5gt^2
The quadratic equation is 2.7 = 7.27t - 4.9t^2
The equation cannot be solve because the value to be square root is a negative.
 
  • #18
Solved. I made a arithmetic error.
 

Related to Throwing a projectile so that it clears the window

1. How do I determine the appropriate angle to throw a projectile to clear a window?

The appropriate angle for throwing a projectile to clear a window can be determined using the equation θ = tan-1(2h/gd), where h is the height of the window, g is the acceleration due to gravity (9.8 m/s2), and d is the horizontal distance between the thrower and the window. This equation can be solved for the angle θ, which will give you the optimal angle to throw the projectile.

2. Does the weight of the projectile affect its ability to clear the window?

Yes, the weight of the projectile does affect its ability to clear the window. Heavier projectiles will require more force to be thrown at the appropriate angle, while lighter projectiles may not have enough force to clear the window. It is important to consider the weight of the projectile when determining the appropriate angle for throwing.

3. Can the height of the thrower affect the trajectory of the projectile?

Yes, the height of the thrower can affect the trajectory of the projectile. The higher the thrower is positioned, the longer the projectile will stay in the air and the farther it will travel. This means that a higher thrower will need to use a smaller angle to clear the window compared to a lower thrower.

4. Is it possible to throw a projectile so that it clears a window at any distance?

Yes, it is possible to throw a projectile so that it clears a window at any distance. However, the angle and force required will vary depending on the height of the window, the distance between the thrower and the window, and the weight of the projectile. As mentioned before, the equation θ = tan-1(2h/gd) can be used to determine the appropriate angle for any distance.

5. Are there any other factors that can affect the trajectory of a projectile thrown to clear a window?

Yes, there are other factors that can affect the trajectory of a projectile thrown to clear a window. These include air resistance, wind speed and direction, and the shape and size of the projectile. These factors may alter the optimal angle and force required for the projectile to clear the window, so it is important to consider them when determining the appropriate angle for throwing.

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