Basketball

[bindeweede]

I think this might be one for Cuddles.

Hi! can anyone help me solving this problem.

A 2.00m tall basketball player is standing on the floor 10.00 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 basketball so that it goes through the hoop without striking the backboard?

The height of the basket is 3.05 m.

[MischiefMonkey]

Having today watched 'Space Jam' for the (approx) 27th time, all I can say is 'What's Up Doc'

[Mulder]

There's a coincidence!

[Matt]

Reminds me of all those marvelous physics and applied maths problems of my past and all the unjustified assumptions required to solve them. First assume a perfectly spherical cow of uniform density.

Mooooo!

[bindeweede]

There's a coincidence!

Yes, I Stumbled across the question on a science-type site, but the answer was not given. Thought some of the mathematicians here might be intrigued.

I wouldn't know where to start.

[bobdezon]

doesnt the weight of the basketball factor into this equation?

[Matt]

doesnt the weight of the basketball factor into this equation?

No the trajectory would be the same as would the intital velocity whatever the weight, an increased weight would require extra force from the basketball player to achieve that velocity.

This is closely related tot he fact that the cannon ball and the pea dropped simultansoely from the tower land at the same time.

[bobdezon]

Ok, I did find out this question had been asked before.

There are two speed component. One is invertical speed and one is horizonal speed

let Vx = horizonal speed
Vy = vertical speed.

Vx = Vcos40
Vy = Vsin40

The displacement of the ball in horizonal direction
x = vt
10 = Vcos40 t

the time it takes the ball to go throught the basket vertically is also the time it takes the ball to go through the ball horizonally.

solve for t
t = 10 / Vcos40

the distance placement of the ball in vertical direction
Xf = .5at^2 + Vt + Xi

we know that t = 10 / Vcos40
3.05 = .5(-9.8) (10 / Vcos40)^2 + Vsin40 (10 / Vcos40) + 2

now just solve:
1.05 = .5(-9.8) (10 / Vcos40)^2 + Vsin40 (10 / Vcos40)
1.05 = -4.9 (10 / Vcos40)^2 + 8.390
-7.34 = -4.9 (10 / Vcos40)^2
1.498 = (10 / Vcos40)^2
1.223 = 10 / Vcos40
1.223 Vcos40 = 10
V = 10 / (1.223 cos40)
V = 10.67 m/s

so the intial velocity of the ball is about 10.67 m/s

first thing is to figure out how long it takes to reach the goal, to do this use the horizontal velocity

vh = v*cos(theta)

then

t = vh/10m

now you know how long it takes to get there

in the vertical direction use the distance equation of motion

s = s0+ vy*t -1/2gt^2

your knowns are

s0 = 0
vy = v * sin(theta)
t = vh/10 = v*cos(theta)/10

plugging them into the equation

s = v * sin(theta) - 1/2 g ( v*cos(theta)/10)^2

s = v*sin(theta) - 1/200 g v^2*cos^2(theta)

we know that cos^2 = 1 - sin^2

substitute in and gather all terms onto one side

s = v*sin(theta) - 1/200 * g * v^2 * (1 - sin^2(theta))

1/200 * g * v^2 * (1 - sin^2(theta)) - v*sin(theta) - s = 0

-1/200 * g * v^2 * sin^2(theta) - v*sin(theta) + (1/200 * g * v^2 - s) = 0

v^2 (-1/200*g*sin^2(theta) + 1/200*g) - v (sin(theta)) - s = 0

everything is known except for "v" so solve using the quadratic equation and you'll get two solutions. use the higher number.

[Mulder]

Of course you also have to consider air resistance and, if outside, wind direction and speed.

[bobdezon]

Yeah I initially figured wind resistance, weight of object thrown and musculature of the the player and the potential energy contained within said muscles. I suppose I was over analysing the situation.

I wonder if the results would be different if the player had previously injested a homeopathic energy booster compound

[Cuddles]

I think this might be one for Cuddles.

I would help, but I'm afraid it mentions the ball going through the hoop, and that never happens when I try.

[ZERO]

Of course you also have to consider air resistance

Ballistic co-effeicient?

[Matt]

Don't forget the spin of the ball. You can aim short but still have the ball roll on with top spin. So you'll need the weight and diameter of the ball, its second moment of inertia and the coefficient of friction between ball and hoop.

Also we've been assuming that the ball starts from a height of 2.0 m and 10.0 metres away from the hoop. Aren't we forgetting that most basket ball players have arms and don't just launch the ball from the top of their head?

[ZERO]

What about the coriolis effect?
Have to allow for the motion of the earth under the ball during flight time.