Problem with projectiles launched horizontally

[Aracon504]

This is a problem which, allegedly requires use of a trigonometric equation (so I've heard, but it is extra credit and thus no help beyond that will be given...)


The problem:

Libyan basketball player Suleiman Nashnush was the tallest basketball player ever. His height was 2.45 meters. Suppose Nashnush throws a basketball horizontally from a level equal to the top of his head. If the speed of the basketball is 12.0 m/s when it lands, what was the ball's initial speed? (Hint: Consider the components of final velocity...)



I'm not really looking for the answer persay, but it is my understanding that I must separate the vertical and horizontal components of the velocity, and I have no idea where to begin without being given an angle of displacement. Any help would be awesome, thanks in advance!

 

[Doc Al]

Yes, treat horizontal and vertical motion separately. (One is accelerated, the other not.) You know the initial angle--he threw it horizontally.

 

[Aracon504]

Yes, treat horizontal and vertical motion separately. (One is accelerated, the other not.) You know the initial angle--he threw it horizontally.

I am completely and utterly confused at this point. What I believe I am sure of.
-The horizontal motion is constant, since we are not taking into account fluid friction.
-The only component being accelerated is thus the vertical velocity, the force accelerating it is gravity...
-The initial velocity is equal to the horizontal velocity in this case, because it is the only mangitude of motion at this point, no?



...So where do I go from here?

 

[Doc Al]

So far, so good. Initially, the motion is purely horizontal, which means the initial velocity in the vertical direction is zero. What will it be after falling to the ground? That will be the vertical component of the final velocity. (You already know the horizontal component.)

 

[Aracon504]

So far, so good. Initially, the motion is purely horizontal, which means the initial velocity in the vertical direction is zero. What will it be after falling to the ground? That will be the vertical component of the final velocity. (You already know the horizontal component.)

I do know the horizontal component?
Given:
-The angle launched is 0 degrees, correct?
-Vertical displacement 2.45m
Final velocity= 12.0 m/s
Again I'm pretty much out of ideas.

 

[Doc Al]

Forget about the horizontal component and concentrate on the vertical. If you drop a ball (initial velocity in the vertical direction = 0), how fast will it be going after falling 2.45m? Depending upon which kinematic equations you have at your disposal, you can solve this is one or two steps.

 

[Aracon504]

final vertical velocity= (-g)(√-2∆y/-g) (This is assuming that the object had no vertical velocity to begin with)(I just derived this equation from two that I know...so correct me if I am wrong)=-6.9 m/s

Pythagorean Thereom= a^2+b^2=c^2
a=√c^2-b^2

=9.8 m/s99% sure this is correct- so correct me if I am wrong...The interesting part is, this was one of the first ways I attempted to solve this problem. It seems that when I enter it (as I reentered it again as soon as I finished) the quantities into my calculator, somewhere along the way I am making a mistake, about 50% of the time, which gave me the incorrect answer...anyhow I appreciate you working with me!

 

[Doc Al]

final vertical velocity= (-g)(√-2∆y/-g)


(I just derived this equation from two that I know...so correct me if I am wrong)


=-6.9 m/s

Good; that's the vertical component of the final velocity.

(I would write the equation as: $v = \sqrt{2 g \Delta y}$.)

Pythagorean Thereom= a^2+b^2=c^2
a=√c^2-b^2

=9.8 m/s

Good. This is the horizontal component, which doesn't change. (Earlier, I had meant to say you could figure out the horizontal component, since it's constant, not that you already knew it. Sorry if that threw you off.)

99% sure this is correct- so correct me if I am wrong...

Looks good to me.

 

[Aracon504]

Good; that's the vertical component of the final velocity.

(I would write the equation as: $v = \sqrt{2 g \Delta y}$.)Good. This is the horizontal component, which doesn't change. (Earlier, I had meant to say you could figure out the horizontal component, since it's constant, not that you already knew it. Sorry if that threw you off.)Looks good to me.

Did not really throw me off, questioned it, but hey that's how it goes. Your equation will probably eliminate the percentage of error I am getting when entering this into my calculator...

What really threw me off, was having my teacher telling me this was an incorrect method of solving it, and that it was extra credit; because it utilized some scarce trig equation...At that point I was probably complicating the problem more than it should have been...

Thanks again!

 

[Doc Al]

Your equation will probably eliminate the percentage of error I am getting when entering this into my calculator...

It's always a good policy to simplify the algebra as much as possible before reaching for the calculator.