Help Brad Understand: Aiming for a Coconut With His Slingshot

[brad sue]

Hi,
Please help me to understand this question of the problem:

A boy wants to knock down a coconut with a rock and his slingshot. He observes that the coconut is about 3.0m above his slingshot and the tree is 4.0m away along the ground.
He knows from experience that the release speed of his rock is 20m/s.
How far above the coconut should he aim?

I don't understand what I am asking to do here. Please can someone help me?

brad

 

[Doc Al]

If gravity didn't exist, the boy would aim directly at the coconut, since the rock would travel in a straight line. But gravity does exist. Realize that, compared to where it would have gone with no gravity, the rock falls a certain distance. That distance is how far above the monkey you have to aim.

Another way to look at it: What angle must the rock be shot at to hit the coconut? Once you find that angle, you can see where the straight line path would have been.

 

[brad sue]

If gravity didn't exist, the boy would aim directly at the coconut, since the rock would travel in a straight line. But gravity does exist. Realize that, compared to where it would have gone with no gravity, the rock falls a certain distance. That distance is how far above the monkey you have to aim.

Another way to look at it: What angle must the rock be shot at to hit the coconut? Once you find that angle, you can see where the straight line path would have been.

Doc, I have a problem to find the angle. I try to use the equations of motion but I have too many unknow : t_final, v_y/final, and the angle itself.

v_y/final=v_y/initial-g(t_y/final)

y_final=y_initial+v_y/initial*t_final-.5*(g)y_final²

those two equations give respectively:

v_y/final=20*sin(θ )-9.81*t_final

4=0+ 20sin(θ )*t_final-4.9*t_final²

Do i miss something here?
please help

brad

 

[Doc Al]

You have the vertical distance as a function of time. Good! But what about the horizontal distance?

 

[brad sue]

You have the vertical distance as a function of time. Good! But what about the horizontal distance?

x_final=x_initial+v_x/initial*t_final
That gives:
3=0+20cos(θ )* t_final

 

[Doc Al]

x_final=x_initial+v_x/initial*t_final
That gives:
3=0+20cos(θ )* t_final

Good! (I think you mean 4, not 3.) Now combine that with the equation for vertical motion.

 

[Doc Al]

Two things:
(1) I think you have mixed up the 3 and the 4. According to your first post:

the coconut is about 3.0m above his slingshot and the tree is 4.0m away along the ground

(2) Here's a trick that may help you combine those two equations. For each equation, isolate the term with the sin or cos. Then square both sides of each equation. Then add them. (I assume you know a useful trig identity about $\sin^2\theta + \cos^2\theta$.)

 

[brad sue]

Two things:
(1) I think you have mixed up the 3 and the 4. According to your first post:


(2) Here's a trick that may help you combine those two equations. For each equation, isolate the term with the sin or cos. Then square both sides of each equation. Then add them. (I assume you know a useful trig identity about $\sin^2\theta + \cos^2\theta$.)

Well I found somthing that is not realistic
I change the 3 and 4 into the right equations and I found t=3.92s

and the angle is .99 degree!

What do you think?

 

[Doc Al]

Well... I didn't crank out the numbers myself, but does your answer make any sense? After all, without gravity the angle would be $\tan \theta = 3/4$. So with gravity, the angle must even be greater. Recheck your calculations. (I'll do it myself when I get a few minutes.)

 

[brad sue]

Well... I didn't crank out the numbers myself, but does your answer make any sense? After all, without gravity the angle would be $\tan \theta = 3/4$. So with gravity, the angle must even be greater. Recheck your calculations. (I'll do it myself when I get a few minutes.)

ok i'll do that

 

[Doc Al]

I did the calculation and found, as expected, that the angle is slightly greater than that needed for straightline motion. (Note: When solving the quadratic equation, there are two solutions. Only one of them is the one we want.) If you still get an unrealistic answer, post the steps just as you did them.

 

[brad sue]

I did the calculation and found, as expected, that the angle is slightly greater than that needed for straightline motion. (Note: When solving the quadratic equation, there are two solutions. Only one of them is the one we want.) If you still get an unrealistic answer, post the steps just as you did them.

I have te two equations:

3=0+ 20sin(θ )*t_final-4.9*t_final²

4=20*cos(θ)*t_final

Isolate the sin and cos

sin(θ)=(3+4.9*t_final²)/20*t_final

cos(θ)=4/20*t_final

I combinethe squares of sin and cos

cos(θ)²+sin(θ)²=(9+29.4*t_final²+24.01*t_final²+16)/(400*t_final²)

equivalent to

1=(25+29.4*t_final²+24.01*t_final⁴)/(400*t_final²)


equivalent

(25+29.4*t_final²+24.01*t_final⁴)=400*t_final²



(25-370.6*t_final²+24.01*t_final⁴)=0

solving this I found 4 values of t but one is correct t=3.92s

Please tell me what is wrong here.

 

[Doc Al]

Your work looks correct. Treat the final equation, as I'm sure you did, as a quadratic in t^2 (say X = t^2). The quadratic has two solutions: you just picked the wrong one! (When you take the square root of those solutions, you can ignore the negative values.)

There are two ways to hit the coconut: The long way or the short way. The long way is essentially shooting it up in the air in tall arc. That's not the one you want.

 

[brad sue]

Your work looks correct. Treat the final equation, as I'm sure you did, as a quadratic in t^2 (say X = t^2). The quadratic has two solutions: you just picked the wrong one! (When you take the square root of those solutions, you can ignore the negative values.)

There are two ways to hit the coconut: The long way or the short way. The long way is essentially shooting it up in the air in tall arc. That's not the one you want.

I tried your suggestions
I checked and rechecked, I found the same thing.
I rechecked the equation too. they sound fine!

 

[Doc Al]

(25-370.6*tfinal2+24.01*tfinal4)=0

solving this I found 4 values of t but one is correct t=3.92s

Please list those four solutions.

Alternatively, if I view this as a quadratic in x = t^2:

$$24.01x^2 - 370.6x + 25 = 0$$

This equation has two solutions. What are they?

 

[brad sue]

Please list those four solutions.

Alternatively, if I view this as a quadratic in x = t^2:

$$24.01x^2 - 370.6x + 25 = 0$$

This equation has two solutions. What are they?

the four solutions I found are:
x=3.9201...
x=0.2602...
x=-0.2602...
x=-3.9201

I get rid of the negative values and I have:

x=3.9201...
x=0.2602... but I only take x=3.9201...

when I use you equation , I have
x=15.36...
x=0.06775...

 

[Doc Al]

I get rid of the negative values and I have:

x=3.9201...
x=0.2602... but I only take x=3.9201...

And why do you ignore the other answer? That's the one you want!

when I use you equation , I have
x=15.36...
x=0.06775...

Right. And when you take the square roots, you get the same four solutions.

 

[lightgrav]

Look, the coconut is 5m away from the start
(along a diagonal). If the stone travels 20m in 1 sec
then (ignoring gravity) it takes about 1/4 sec to go 5m.

WITH gravity, in 1/4 sec the stone deviates from its path
by 0.3 m, so you have to aim about twice as high,
and stone will take anout twice as long ... about half sec.

Why do you inisist on discarding the t^2 = 0.2602 [s^2]
(which means t about .51 sec) ?
You know that during 2 seconds of free-fall,
a stone would deviate from its path by nearly 20 meters
- so you want to aim 20 meters high?
That's what DocAl meant by "the long way" almost straight up.