How to Calculate Initial Velocity for a Protein Bar to Reach a Hiker on a Slope?

[negation]

Homework Statement

You toss a protein bar to your hiking companion located 8.6m up at 39° slope. Determine the initial velocity vector so the bar reaches your friend moving horizontally.

The Attempt at a Solution

assuming the projectile lands on the friend at the apex of the trajectory

8.6 cos 39° = 6.683m = x
8.6 sin 39° = 5.41215m = y

vyf^2 - vyi^2 = 2g(yf - yi)
vyi = 10.299ms^-1
vyi = vi sin 39° = 10.299
vi = 10.299/sin 39° = 16.365ms^-1

 

[BvU]

What is your question ?

 

[negation]

What is your question ?

Determine the initial velocity vector so the bar reaches your friend moving horizontally.
Is my solutions valid?

 

[BvU]

Velocity is a vector. you have the right vyi
For the magnitude you assume you throw straight at your friend. That is a bit worrying. The trajectory is not a straight line, so the bar might just hit the ground before reaching your friend. Either that, or he has to stoop.
Better assume the throwing and catching takes place at shoulder height for both.

If I were you in the exercise, I would throw at an angle > 39 degrees!

For the horizontal component, you have a simple uniform motion.
make a drawing (I take it you already did?).

Have to go. Good luck!

 

[negation]

Velocity is a vector. you have the right vyi
For the magnitude you assume you throw straight at your friend. That is a bit worrying. The trajectory is not a straight line, so the bar might just hit the ground before reaching your friend. Either that, or he has to stoop.
Better assume the throwing and catching takes place at shoulder height for both.

If I were you in the exercise, I would throw at an angle > 39 degrees!

For the horizontal component, you have a simple uniform motion.
make a drawing (I take it you already did?).

Have to go. Good luck!

I understand it's not a straight line due to the effect of gravity. If gravity were absence, then a straight line would be truism.
I'll think about it

 

[negation]

Since the projectile reaches the companion at the apex of it's trajectory,
then, t = vyi/g.
total time = 2vi sin (a) /g
x = vi cos (a). t = vi cos (a) [2vi sin(a) /g] = [ vi^2 sin(2a) ]/g = 8.6

Unknown variable a and vi, how should I go about?

 

[lightgrav]

I thought x = 6.683 m ... and if you find time (-vy1/g) , you get vx easily ... (it is constant)

 

[BvU]

I excelled a little to draw what we had so far in post #1 (vi = 16.4 m/s, 39 degrees wrt horizontal).
As you see, the bar would be moving horizontally at the y level of your buddy if there were no slope. This is at t=vyi/g, by which time x is already over 13 m.

In post #6 you continue with

total time = 2vi sin (a) /g

which comes out of the blue as far as I can see. In other situations a 2 comes in because we can use time from ground to apex = time from apex to ground, but that is not what we have here: catch is at the apex!

So the thing has to be thrown a bit higher up, as you already indicate in your drawing.
Bravo for doing the drawing. But: we were making such good progress in the x-y coordinate system in which g points in the -y direction. Much more comfortable to stay in that system a little more, and look for the angle to throw wrt horizontal. When I look a little longer, I see a theta = a, so I take it you mean wrt horizontal the angle is 39 + a. OK. And the vector pointing to the upper right is v_i
But the expressions are not correct: vi cos (a+39) should point horizontally to the right and vi sin (a+39) is the y-component, pointing straight up. Also, the curved line can't be right: the bar never ever has a negative vy.


Remember, vyi has already been established and is correct: we can't change it any more. It's the only way to get vy = 0 at the 5.4 meter higher where friend is. Corresponding time is indeed vyi/g, 1.05 sec. That too can't be changed any more! And yes, friend is at the apex of the bar's trajectory in the air.

Now all we have to do is aim a little higher and at the same time throw a little less hard in such a way that vyi stays the same. Only the horizontal speed has to change. How much ? Well, so much that the bar is at the x-position of our dear friend at the moment it is moving horizontally (which as we agreed is at time vyi/g).

So we have one equation with one unknown. Solve and calculate vi as well as the angle wrt horizontal !

 

[negation]

I excelled a little to draw what we had so far in post #1 (vi = 16.4 m/s, 39 degrees wrt horizontal).
As you see, the bar would be moving horizontally at the y level of your buddy if there were no slope. This is at t=vyi/g, by which time x is already over 13 m.

In post #6 you continue with which comes out of the blue as far as I can see. In other situations a 2 comes in because we can use time from ground to apex = time from apex to ground, but that is not what we have here: catch is at the apex!

So the thing has to be thrown a bit higher up, as you already indicate in your drawing.
Bravo for doing the drawing. But: we were making such good progress in the x-y coordinate system in which g points in the -y direction. Much more comfortable to stay in that system a little more, and look for the angle to throw wrt horizontal. When I look a little longer, I see a theta = a, so I take it you mean wrt horizontal the angle is 39 + a. OK. And the vector pointing to the upper right is v_i
But the expressions are not correct: vi cos (a+39) should point horizontally to the right and vi sin (a+39) is the y-component, pointing straight up. Also, the curved line can't be right: the bar never ever has a negative vy.Remember, vyi has already been established and is correct: we can't change it any more. It's the only way to get vy = 0 at the 5.4 meter higher where friend is. Corresponding time is indeed vyi/g, 1.05 sec. That too can't be changed any more! And yes, friend is at the apex of the bar's trajectory in the air.

Now all we have to do is aim a little higher and at the same time throw a little less hard in such a way that vyi stays the same. Only the horizontal speed has to change. How much ? Well, so much that the bar is at the x-position of our dear friend at the moment it is moving horizontally (which as we agreed is at time vyi/g).

So we have one equation with one unknown. Solve and calculate vi as well as the angle wrt horizontal !

I have solved the question earlier in the day. Apparently, I overlooked one of the variable the answer of which I have already found. Still, Thanks for the effort.

I'll post the solution up just in case someone requires the same help in the future.