What is the Initial Speed of a Basketball Thrown to a Hoop?

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The discussion focuses on calculating the initial speed of a basketball thrown at a 40-degree angle from a height of 2.00m to a hoop 3.05m high, located 10.0m away. The initial approach using energy equations is deemed less effective than applying kinematic equations of motion. Participants suggest using the kinematic equations to determine the time it takes for the ball to reach the hoop and then solving for the initial velocity. The conversation emphasizes setting up simultaneous equations for both x and y components to find the solution. The use of standard equations is encouraged for clarity and accuracy in solving the problem.
eatblue
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Homework Statement


The problem

A basketball player 2.00m in height throws the ball at a 40degree angle 10.0m from the basketball post and manages to get the ball through the net without it touching the backboard. If the height of the net is 3.05m from the ground, what is the initial speed of the ball?


Homework Equations



(PE+KE)initial = (PE+KE)final
PE = mgh, KE = (1/2)mv2
W = (Fcosθ)Δx , θ is the angle between the directions F and Δx

The Attempt at a Solution


What I tried:

(PE+KE)initial = (PE+KE)final
set y=0 at the position the ball was thrown
0 + .5vi2 = gh + .5vf2
-> .5vi2 = 9.80m/s2*1.05m + .5vf2

I have no idea how it should be handled from here and this is all the information that was given...
 
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eatblue said:

Homework Statement


The problem

A basketball player 2.00m in height throws the ball at a 40degree angle 10.0m from the basketball post and manages to get the ball through the net without it touching the backboard. If the height of the net is 3.05m from the ground, what is the initial speed of the ball?


Homework Equations



(PE+KE)initial = (PE+KE)final
PE = mgh, KE = (1/2)mv2
W = (Fcosθ)Δx , θ is the angle between the directions F and Δx

The Attempt at a Solution


What I tried:

(PE+KE)initial = (PE+KE)final
set y=0 at the position the ball was thrown
0 + .5vi2 = gh + .5vf2
-> .5vi2 = 9.80m/s2*1.05m + .5vf2

I have no idea how it should be handled from here and this is all the information that was given...


Welcome to the PF. I don't think that using the energy equations is the easiest way to do this problem. This is pretty much a straightforward application of the kinematic equations of motion. Start with the kinematic equations:

y(t) = y(0) + _____ + _____ (fill in the starting equations)

x(t) = x(0) + ______ (there are on ly two terms here -- why?)

Vy(t) =

Vx(t) =

And write the acceleration formulas for x and y..

Then the usual way to solve this type of problem is use the x(t) and Vx(t) information to solve for the time it takes for the ball to reach the hoop, and use that time answer to finish solving the problem.

In the case of this question, you may end up with a couple simultaneous equations (some in x, some in y) that you solve together to get the initial velocity of the ball.

Give it a go with the standard equations. If you have trouble still, post your work here for folks to try to help you out.
 
Use y(t) = -4.9t2+v0t+y0
and x(t) = v0t+x0?
Thanks! I'd completely forgotten about that formula:)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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