Projectile Motion: Using an arrow to shoot a coconut dropped by a monkey

[songoku]

Homework Statement

A person on ground wants to shoot coconut hold by monkey sitting on tree at certain height from the ground. He shoots the arrow at angle 35 degree with respect to horizontal at the same time the monkey drops the coconut. Given that the arrow hits the coconut 1 second after the shot, find (in any order) the initial speed of arrow and the height of monkey from ground

Relevant Equations

Projectile Motion

Let:
$x$ = horizontal distance from person to tree
$h_t$ = height where arrow hits the coconut (measured from ground)
$h_o$ = height of monkey
$u$ = initial speed of arrow
$\theta = 35^0$
$t = 1~ \text{s}$

Vertical displacement of coconut until it is hit:
$$h_t - h_o = -\frac{1}{2} g t^2 = -\frac{1}{2} (9.81) (1^2)=-4.905...(1)$$

Horizontal distance traveled by arrow:
$$x=u ~\text{cos} ~\theta . t=u ~\text{cos} ~35^0...(2)$$

Height of arrow when hits coconut:
$$h_t = u~ \text{sin} ~\theta . t - \frac{1}{2}gt^2$$
$$h_t=u~ \text{sin} ~35^0 - 4.905...(3)$$

Substitute (3) to (1):
$$u ~\text{sin} ~35^0-4.905-h_o=-4.905$$
$$u ~\text{sin} ~35^0=h_o$$

Then I am stuck.

Thanks

 

[haruspex]

You have not used the fact that the arrow is shot from ground level.

 

[songoku]

You have not used the fact that the arrow is shot from ground level.

I think I have used that in this part:

Height of arrow when hits coconut:
$$h_t = u~ \text{sin} ~\theta . t - \frac{1}{2}gt^2$$
$$h_t=u~ \text{sin} ~35^0 - 4.905...(3)$$

The LHS should be: final height of arrow - initial height of arrow and because initial height = 0 the LHS is only final height of arrow

Or maybe you mean I have to use that fact on other part of my working?

Thanks

 

[haruspex]

I think I have used that in this part:

Sorry, you're right.

Ok. so suppose we have some solution. What happens if we double x? We would have to double u to keep the time the same, and double h_o. h_o-h_t would remain the same. And that's it; all the given data still applies.

So you do not have enough info.

 

[songoku]

Thank you very much haruspex

 

[PeroK]

Thank you very much haruspex

I wonder whether the time taken to hit the coconut is the same as it would be if there were no gravity? If the arrow moved in a straight line and the coconut didn't fall?

 

[haruspex]

I wonder whether the time taken to hit the coconut is the same as it would be if there were no gravity? If the arrow moved in a straight line and the coconut didn't fall?

Depends what you allow to vary. It's fixed by u, θ, x.

 

[sysprog]

I wonder whether the time taken to hit the coconut is the same as it would be if there were no gravity? If the arrow moved in a straight line and the coconut didn't fall?

I think that the time taken would not be the same, because the straight line path is shorter than the curved trajectory, right?

 

[haruspex]

I think that the time taken would not be the same, because the straight line path is shorter than the curved trajectory, right?

If the horizontal distance and horizontal velocity are unchanged then the time is the same.

 

[sysprog]

If the horizontal distance and horizontal velocity are unchanged then the time is the same.

How would the horizontal velocity be unchanged, if the archer has to aim at a non-horizontal upward angle when there is gravity, and can aim along a non-arched exactly horizontal path when there is no gravity?

 

[PeroK]

How would the horizontal velocity be unchanged, if the archer has to aim at a non-horizontal upward angle when there is gravity, and can aim along a non-arched exactly horizontal path when there is no gravity?

If the solution is independent of $g$, then we could set $g = 0$.

 

[sysprog]

If the solution is independent of $g$, then we could set $g = 0$.

Wouldn't that eliminate the necessity of the curve, and thereby reduce the length of the path along which the arrow travels, and consequently reduce the time it takes for it to reach the target?

 

[haruspex]

How would the horizontal velocity be unchanged, if the archer has to aim at a non-horizontal upward angle when there is gravity, and can aim along a non-arched exactly horizontal path when there is no gravity?

There is no horizontal trajectory option.
The coconut starts at elevation angle θ above above the archer, i.e. at height $x\sin(\theta)$. The arrow is aimed at the same elevation. The coconut is released at the instant the arrow is released, speed u.
Without gravity, the coconut doesn't move and the arrow travels in a straight line to hit it.
With gravity, the arrow travels in a parabola and hits it.
In both cases, the time taken is $\frac{x}{u\cos(\theta)}$.

 

[sysprog]

There is no horizontal trajectory option.

Let's please suppose that there were such an option $-$ the archer's boss has given him a 'gravity on-off' button, and with no gravity, our archer presumably would find it easy enough to 'ascend' a parallel tree.

Now, supposing that he is in the tree, and at 180 degrees (exactly horizontal) to the target, he has the option to use the button to activate the 'gravity on' or gravity off' condition:

If the target is at such a great distance that, with the 'gravity on' condition active, the arrow will not reach it, unless the archer aims at a 45 degree upward angle (so that (by the law of cosines) gravity is active over only 70.71% of the horizontal distance) and he can then still hit the target; however, with gravity off, distance is not a concern, and the archer can aim at 0 (in this context otherwise called minus 180) degrees $-$ wouldn't the arrow reach the target sooner with 'gravity off' active?

 

[PeroK]

Let's please suppose that there were such an option $-$ the archer's boss has given him a 'gravity on-off' button, and with no gravity, our archer presumably would find it easy enough to 'ascend' a parallel tree.

Now, supposing that he is in the tree, and at 180 degrees to the target, he has the option to use the button to activate the 'gravity on' or gravity off' condition:

If the target is at such a great distance that, with the 'gravity on' condition active, the arrow will not reach it, unless the archer aims at a 45 degree upward angle (so that (by the law of cosines) gravity is active over only 70.71% of the horizontal distance) and he can then still hit the target; however, with gravity off, distance is not a concern, and the archer can aim at 0 (in this context otherwise called minus 180) degrees $-$ wouldn't the arrow reach the target sooner with 'gravity off' active?

Where do you aim to hit the coconut? Should $\theta = \alpha$, where $\theta$ is the angle of the projectile and $\alpha$ is the angle of elevation of the coconut?

Do the calculations, please.

Does the answer depend on $g$? What if $g = 5m/s^2$ or $g = 1m/s^2$ or $g = 0.00001m/s^2$.

Do the calculations please.

Can you study the problem in an accelerating reference frame, falling at $g$? In that frame do you have straight-line motion that results in the same answer as above?

PS if there is a possibility of the coconut hitting the ground before the arrow reaches it, then we need to factor that into the constraints of the problem.

 

[jbriggs444]

So you do not have enough info.

As I read the problem, we are are expected to treat the initial height of the monkey ($h_0$ using OP's convention) above the ground as a known input. We would then be expected to produce a result written in terms of $h_0$.

Edit: But that can't be right as @kuruman points out. The problem statement asks for a calculation of $h_0$.

@PeroK has been giving strong hints about the relevance of gravity to the problem.

 

[kuruman]

As I read the problem, we are are expected to treat the initial height of the monkey (h0 using OP's convention) above the ground as a known input. We would then be expected to produce a result written in terms of h0.

And since the problem is asking to "find (in any order) the initial speed of arrow and the height of monkey from ground", OP has first "found" the height to be $h_0$ and must now look for the speed. Or is it that OP found the speed first and must now look for the height? OP's equation $u ~\text{sin} ~35^0=h_o$ in post #1 says it all.

 

[haruspex]

Let's please suppose that there were such an option $-$ the archer's boss has given him a 'gravity on-off' button, and with no gravity, our archer presumably would find it easy enough to 'ascend' a parallel tree.

Then this has nothing to do with the context of your post #8 or my post #9. I see no benefit in discussing it further.