Classwork Problem (please don't move to homework!)

[grayfalcon89]

Hi,

 

My physics class just covered projectile motion and this problem was like the at the end of lesson problem. We get a bonus credit if we get it right! I got the answer but I want to know if this is right (I got all the other projectile motion questions right so far though).

 

* A basketball player tries to make a half-court jump shot, releasing the ball at the height of the basket. Assuming that the ball is launched at 51.0 degrees, 14.0 m from the basket, what speed must the player give the ball?

 

My Solution:

 

Let the resultant vector be v. Then the horizontal component velocity vector is v*cos 51 and vertical is v*sin 51.

 

Focusing on the vertical side:

 

vo = v sin 51

a = 9.80 m/s^2

d = vo*t+1/2*a*t^2

derivative gives vf = vo + at

Since coming back velocity is same as starting except opposite:

- v sin 51 = v sin 51 + (-9.80)t

- 2v sin 51 = -9.80t

t = (10v sin 51)/(49)

 

Now, horizontal:

 

d = vt

14 = (v cos 51)(10 sin 51 * 1/49)

Solving for v gives me 11.8 m/s.

[swansont]

You dropped a "v" from your last equation.

 

Substitute your numbers back in to an equation and see if it holds.

 

v_f = v_o + at

 

Put in v_o, a and t and see if you get -v_o

 

and why shouldn't this be in homework?

[grayfalcon89]

What are you talking about?

 

I clearly stated that horizontal component vector is v*cos 51. This is the "v." And as for "v" in the equation, I took the square root to get v = 11.8 m/s.

 

And no, I said that this should not belong to homework because I'm not posting a homework and asking you to give my answers. I'm asking guide on this classwork problem that I'm not sure of and even posted what I believe is answer. It's more rational to say that it is a question about concept than simple assertion of homework.

[Cap'n Refsmmat]

Your answer checks out, I think. Running it back through the same equations gives me a similar answer.

[Phi for All]

And no, I said that this should not belong to homework because I'm not posting a homework and asking you to give my answers. I'm asking guide on this classwork problem that I'm not sure of and even posted what I believe is answer. It's more rational to say that it is a question about concept than simple assertion of homework.

We do *not* give you your answers in Homework Help. Asking guidance on a classwork problem and receiving helpful hints and nudges towards the answer is what that section is all about.

 

That's why we move obvious homework questions there. The other sub-forum posters are much more likely to just give you answers and that doesn't really help students in the long run.

[swansont]

What are you talking about?

 

I clearly stated that horizontal component vector is v*cos 51. This is the "v." And as for "v" in the equation, I took the square root to get v = 11.8 m/s.

 

You took a square root, yet there is only one "v" in the equation you posted. It's missing a "v"

 

14 = (v cos 51)(10 sin 51 * 1/49) should actually be

14 = (v cos 51)(10 v sin 51 * 1/49)

 

And yes, it checks out, as the Cap'n confirms. What I suggested was a way you could check it out for yourself.

vf = vo + at

and you know that vf = -vo

if your values are right, the equation will be satisfied.

[xptoast]

Trick question? There was no weight for the ball as it can only be implied from the statement you first made. It seems there may be other missing variables as the real world needs weight and the force of the guys throwing arm and such. I don't know, just guessing.

[Bignose]

Trick question? There was no weight for the ball as it can only be implied from the statement you first made. It seems there may be other missing variables as the real world needs weight and the force of the guys throwing arm and such. I don't know, just guessing.

 

No. The mass (weight) doesn't matter. Assuming it remains constant (and there is no reason not to assume this in this problem) it drops out of the equation.

 

Drop a bowling ball and a tennis ball from the same height and they will hit the ground pretty much simultaneously. One can only say "pretty much" because the different balls will experience different drags, but the results are quite close. Once you can remove the atmospheric drag, any two objects can be dropped at the same time and hit the ground at the same time. The Apollo astronauts proved that by dropping a feather and a hammer at the same time on the Moon and they hit the ground at the same time.

 

Please see this link http://er.jsc.nasa.gov/seh/feather.html