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posted by  A.G on 9/13/2008 11:22:51 PM  |  status: Live  

physics help ! please

Course Textbook Chapter Problem
General Physics N/A N/A N/A
Question Details:

 

A basketball player 2.01 m tall wants to make

a basket from a distance of 10.9 m. The hoop

is at a height of 3.05 m.

The acceleration of gravity is 9.8 m/s2 .

Neglect air friction.



If he shoots the ball (from a height of

2.01 m) at a 24.3 ? angle, at what initial

speed must he throw the basketball so that

it goes through the hoop without striking the

backboard? Answer in units of m/s.
 
 
 
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Apprentice
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posted by TCCKHOA on 9/14/2008 12:31:27 AM  |  status: Live
Asker's Rating: Helpful   
A.G 's comment:
"thank you"
Response Details:
           Obviously, this is the projectile problem.
           Let the positive direction be upward. Let the origin right at the point the player thows the ball.
 
 
 
 
We must have  when
 
            (1)
 
                       (2)
 
Substituting (1) into (2) we have:
 
 
 
  After some calculations, we get the final answer is  . We choose the positive value in this problerm.
 
 I hope that I do all the calculations correctly.
 
 
 
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posted by Jack Alope on 9/14/2008 12:38:43 AM  |  status: Live
Asker's Rating: Lifesaver   
A.G 's comment:
"very good "
Response Details:

Find initial velocity:
Vx = cos(24.3)V = 0.9114V
Vy = sin(24.3)V = 0.4115V

Time:
x = vt
10.9m= (cos24.3V)t
t = (10.9)/(cos24.3V)

Solve for V:
yf = (1/2)at2 +Vit + yi
3.05m = (1/2)(-9.8)[(10.9)/(cos24.3V)]2 + (sin24.3V)[(10.9)/(cos24.3V)] + 2.01m
1.04 = (-4.9)[(10.9)/(cos24.3V)]2 + 4.92
-3.88 = (-4.9)[(10.9)/(cos24.3V)]2
0.7921 = [(10.9)/(cos24.3V)]2
0.89 = (10.9)/(cos24.3V)
V = (10.9) / [(0.89)(cos24.3)] = 13.42 m/s
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posted by skmjee on 9/14/2008 12:43:33 AM  |  status: Live
Asker's Rating: Helpful   
A.G 's comment:
"thank you"
Response Details:
vo is the velocity of projection
vertical component of the projection = uv =  voSinθ
Horizontal component of projection = uh = voCosθ
let t be the time of flight.
So, uht = 10.9  ----------(1)   [as the ball moves a horiz displacement 10.9 m]
Again the vertical displacement is H(say)
Thus,
H = uvt - 1/2 gt2
1.04 = uvt - 4.9t2                           [as the vertical displacement is H = 3.05 - 2.01 = 1.04 m]
=>uvt-4.9t2 = 1.04     -----------------(2)
Deviding (2) by (1) we get,
(uvt - 4.9t2)/uht = 1.04/10.9
=>(uv - 4.9t)/uh = 1.04/10.9
=> t = (10.9uv - 1.04uh)/(10.9*4.9)
 
Putting t in equation (1) we get,
uh*(10.9uv - 1.04uh)/(10.9*4.9) = 10.9
=>uh*(10.9uv - 1.04uh)=10.9*10.9*4.9
=>voCosθ(10.9voSinθ - 1.04voCosθ)   =   10.9*10.9*4.9
=>vo2(10.9Sin24.3oCos24.3o - 1.04Cos224.3o) = 10.9*10.9*4.9
=>vo2(4.088 - 0.864) = 10.9*10.9*4.9
=> vo2 *3.224 = 582.169
=> vo2  = 582.169/3.224
=> vo2 = 180.574
=> vo = 13.44 m/s
So initia speed of the ball is 13.44 m/s
 
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