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Projectile motion, with no Initial Velocity 
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#1
Sep1608, 04:43 PM

P: 9

1. The problem statement, all variables and given/known data
A 2.00m tall basketball player is standing on the floor 10.0 m from the basket. If he shoots the ball at a 40 degree angle from the horizontal, at what initial speed must he throw the ball so that it goes through the basket without striking the backboard? The height of the basket is 3.05 m. 2. Relevant equations Y=Vi t + 1/2 g t^2 V^2 = + 2 g Y Those are my best guesses, but I'm not entirely sure. 3. The attempt at a solution I don't even know where to begin. 


#2
Sep1608, 04:58 PM

P: 117

First, establish a rectangualr coordinate system, with the origin at say 2m above the floor and 10m from the hoop.
Next, list the known variables and their values. Initial position: (xi, yi) = (0,2) Final position: (xf, yf) = (10, 1.05) ...note that yf is 1.05 m above the origin defined above. Launch angle is 40 deg above horizontal. To find the initial speed, we find the components of the initial velocity, whcih we write Vix = Vi cos 40 Viy = Vi sin 40 Now write the two equations which give the final position (x,y) as a function of initial position, initial speed components and time. You should find that the only unknowns are the initial speed Vi and time of flight to the hoop. Two eqns, two unknowns... 


#3
Sep1608, 05:12 PM

P: 9

I don't really understand what exactly to do when you say to write the equations as a function in this case.



#4
Sep1608, 05:26 PM

P: 9

Projectile motion, with no Initial Velocity
Is there anyone that can further help? I'm very crunched for time tonight and any help would be very appreciated.



#5
Sep1608, 07:20 PM

HW Helper
P: 5,341

For instance, you know that the horizontal Velocity = V*Cosθ If T is the total time of flight, and the basket is 10m away then you know that V*Cosθ*T = 10m Believe it or not you are a good part of the way there already. OK, what else do you know? How much time to maximum height? Initial velocity/g = V*Sinθ/g = T1 Now T1 is only part of the problem, because Total time = Time to rise (T1) plus time to fall (T2). OK so how much time for it to drop from max height to the height of the basket? Figure Max height. Not that hard because (V*Sinθ)^{2} = 2gH (Remember H is 2m higher than the ground.) Now last piece of the puzzle: How long to drop from max height? (H  1.02) = 1/2 g* T2^{2} (The 1.02 is the difference in height above the release point.) Now start solving. Your answer should grind out the bottom. 


#6
Nov409, 10:35 PM

P: 1

1.05=v^2sin40*(10/v^2cos40)0.5*9.8*(10/v^2cos40)^2
1.05=10tan40(490/v^2(cos40)^2) v^2((cos40)^2*10tan40)490=v^2((cos40)^2) v^2((cos40)^2*10tan40)v^2((cos40)^2)=490 v^2(4.337215)=490 v=(490/4.337215)^0.5 v=10.66m/s HOPE YOU UNDERSTAND NOW! 


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