I need help with mah physics

nofx

nofx

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I think theres a simple way to solve this but I cannot think of it. Its a physics problem I wouldve tried pming someone but none of my classmates know and Im not sure who here is good at physics, or math in general.

a 2.00m tall basketball player wants to make a basket from a distance of 10.0m. if he shoots the ball at a 45 degree angle, at what initial speed must he throw the basketball so that it gos through the hoop without striking the backboard.
 
All I can remember from 3rd year physics is V=d/t.

This problem sounds more like accelaration to me, but then again, I was never good at physics.
 
nofx said:
a 2.00m tall basketball player wants to make a basket from a distance of 10.0m. if he shoots the ball at a 45 degree angle, at what initial speed must he throw the basketball so that it gos through the hoop without striking the backboard.
Click to expand...
omgwtf

What year are you in?
 
wouldn't you need to know how high the basketball goal is? ...unless we're assuming it's a standard 10ft goal?

and as my phsyics teacher told me, "All you need to know is f=ma and you can't push with a rope, and you can figure the rest out."
 
The height the basketball goal is 3.05m

wtf i cant believe i didnt include that -.-

*i copied the question word for word from the book and it includes a picture but in the written answer it doesnt state the height but in the picture it does.
 
Basically, you're going to need to split the trajectory of the basketball into x and y components. The velocity of the x component will remain constant, which the y component will move with constant acceleration due to gravity.

Have you had Calculus yet?
 
nacrolepsy:
no i havent had calculus im currently in precalculus which is what you take before you get into actual calculus.. im a sophmore. i already know how to split the voy and vox and i know that vo will be reduced by .7 because of the 45, sin and cos. but i just dont know how to apply tis to the actual problem?

xcellerate

whats your reasoning?
 
use,

s=ut1/2at^2
v^2=u^2+2as
v=u+at

u= initial velocity v= final velocity s=displacement a=acceleration t=time
you need to use those three, maybe f=ma

following on,from thingy^

u need to use a vector diagram.

, look in your text book,,,,,

using one of those equations you need to find the initial velocity = u

s = 10
u = ?
a = -9.81

cant remeber anymore...

io"think i came up with 12.2... i dropped physaics so i dunno if im right or not :P
hate ****ing forces motion....:p
 
Joims said:
use,

s=ut1/2at^2
v^2=u^2+2as
v=u+at

you need to use those three, maybe f=ma
Click to expand...

Those 3 formula's excluding the force mass acceleration formula, were the onces that sprung to mind when I first read the question. Should be able to work it out with those, I seem to remember similar questions and using those formula's. I find that if you list all teh variables you have to start with, you can then select the best formula to use to work out the unknowns.
 
^^^^^^^^^^^^

Really nofx you should be able to use the textbook/internet to figure this one out:)
 
i think you have to use pythagoras's apothem on this one.

c² = √a² + b² (to determine the hypotenuse)

Code: 
  |\
  | \          
a |  \  c       
  |   \  
  |____\
    b

you already got distance from player to basket (b), and the height (a), and your angle (45).

or was it the pythagoras's theorem? the one where you use tan, hip, and sen formulas? :\

... knowing that, i don't know what to do next. try the above tips
 
Narcolepsy said:
Basically, you're going to need to split the trajectory of the basketball into x and y components. The velocity of the x component will remain constant, which the y component will move with constant acceleration due to gravity.

Have you had Calculus yet?
Click to expand...

You do not need calculus to solve this. It's simple algebra and trig.
 
thanks ill try the v^2 = u^2 +2as one...
i didnt work them out but they look kind of dimensional nonsensey
 
nofx said:
a 2.00m tall basketball player wants to make a basket from a distance of 10.0m. if he shoots the ball at a 45 degree angle, at what initial speed must he throw the basketball so that it gos through the hoop without striking the backboard.
Click to expand...

I'm going to assume that this is a standard 3m high basketball net and that the angle and velocity is measured from the ball leaving the player at his head height (2.00m) Now break it into vertical and horizontal.

sin45 and cos45 are the ratios between component and velocity. They are both equal to 2^(-1/2), one over root 2.

So first the horizontal:
Vx=2^(-1/2)*V
10m=Vx*t
t=10m/(2^(1/2)*V)

Then the vertical:
h=Vy*t-1/2gt^2 (h=height, g=gravitation const=9.8m/s^2)
so
1m=2^(1/2)*V*[10m/(2^(1/2)*V)]-4.6m/s^2*[10m/(2^(1/2)*V)]^2
1m=10m-4.6m/s^2*100/(2*V^2)
9m=980m/s^2/V^2
9V^2=980m/s^2
V^2=980/9
V=(980/9)^(1/2)
approx 10m/s
 
Nat Turner said:
You do not need calculus to solve this. It's simple algebra and trig.
Click to expand...
Technically you don't need it, but understanding differentiation and integration can help you understand the relationship between position, velocity, and acceleration. And problems are always easier if you understand exactly what's going on.

i think you have to use pythagoras's apothem on this one.

c² = √a² + b² (to determine the hypotenuse)

Code:

|\ | \ a | \ c | \ |____\ b


you already got distance from player to basket (b), and the height (a), and your angle (45).

or was it the pythagoras's theorem? the one where you use tan, hip, and sen formulas? :\

... knowing that, i don't know what to do next. try the above tips
Click to expand...
Nah, the basketball moves in an arc, so that wouldn't work.
 
So first the horizontal:
Vx=2^(-1/2)*V
10m=Vx*t
t=10m/(2^(1/2)*V)

Then the vertical:
h=Vy*t-1/2gt^2 (h=height, g=gravitation const=9.8m/s^2)
so
1m=2^(1/2)*V*[10m/(2^(1/2)*V)]-4.6m/s^2*[10m/(2^(1/2)*V)]^2
1m=10m-4.6m/s^2*100/(2*V^2)
9m=980m/s^2/V^2
9V^2=980m/s^2
V^2=980/9
V=(980/9)^(1/2)
approx 10m/s
Click to expand...
I roughly understood nothing of that.
 
yeah technically to get those equations I used you are differentiating position to get velocity and then differentiating that to get acceleration, but mostly it's common sense. 1st, position=velocity*time pretty easy. The next one is that position=velocity*time+acceleration/2*time^2. Seems kind of tougher, but acceleration*time is the amount of velocity you've gained right. So you are really saying that position is the average of the initial velocity and final velocity * time. position = time(velocity+acceleration*time/2)
 
Just toss a ball in the air.. it's pretty obvious how it works.
 
Well, technically yes you do need calculus, but if you don't know calculus you can use equations that already have the calculus built in for you. Like x = xo + vo t + 1/2 a t^2.
 
Ok I guess that looks pretty complicated, but only because I had to use ascii to represent square roots. I'll try to explain again:

If you look at your initial velocity you can break it into two components, vertical and horizontal. Because the angle is 45, both of these components are the same, you can draw out the vector and see. It's a simple 45 degree right angle triangle. If the hypoteneuse is 1, then the sides are all 1/sqrt2. So if the hypoteneuse is your velocity (V), then the sides are V/sqrt2.

So we know that the ball is travelling V/sqrt2 m/s in the horizontal axis. We will call the horizontal speed Vx. It has to go 10m. distance=speed*time

10m = Vx*time

we'll isolate time, because we don't actually care what time is.

time = 10/Vx
time=10*sqrt2/Vx
(this will come in handy once we look at gravity)

Gravity accelerates everything down at 9.8m/s^2 whenever you are near the surface of the earth. It's just one of those constants you memorize. If we exclude air resistance, then gravity is the only force. So acceleration = 9.8m/s^2

What I was trying to explain earlier, is that when we throw linear acceleration into the mix, you have a different equation for position. Now there is no constant velocity because the velocity is linearly increasing/decreasing over time. So because it's linear we don't need to bother with messy integrals, we can just take the average of the initial and final velocity and use that.

In this case, the initial velocity is the vertical component(Vy) that the ball is thrown at. The final velocity is the vertical component minus the acceleration * time. This is (Vy-9.8m/s^2*time) So in equation form

Vavg = (Vy+(Vy-9.8m/s^2*time))/2
(the 9.8 is negative because gravity is acting downwards, opposite the initial velocity)
now position, or height is equal to the average velocity * time
height = Vavg*time
height = (Vy+Vy-9.8m/s^2*time)/2*time
(reorganize that a bit)
height = Vy*time/2 + Vy*time/2 - 9.8m/s^2*time*time/2
height = Vy*time - 4.9m/s^2*time^2

If the player is 2m tall and the net is 3m tall, then the ball has to achieve a final height of 1m (relative to where it starts).

I put in 1m for height and 10m/Vx for time
1m = Vy*10m/Vx - 4.9m/s^2*(10m/Vx)^2
(remember that Vy and Vx have the same value and that is V/sqrt2)
1m = 10m - 4.9m/s^2*(10m*sqrt2/V)^2
1m = 10m - 4.9*100m^2*2/V^2
9m = 4.9m/s^2*200m^2/V^2
9m*V^2=980m^3/s^2
V^2=980/9m^2/s^2
V=sqrt(980/9)m/s

V=approx 10.4m/s

This is all based on the assumption that the ball is launched from a height of 2m and the net is 3m high

It takes longer to write everything out when you try to explain it really simply. The actual equations take about a minute to do. I study mechanical engineering, but this is all from highschool.
 
Beerdude26 said:
Rocket science it is
Click to expand...

No, rockets travel in three dimensional space, encounter wind resistance, have limits on fuel capacity, etc.
 
Dan said:
Ok I guess that looks pretty complicated, but only because I had to use ascii to represent square roots. I'll try to explain again:

If you look at your initial velocity you can break it into two components, vertical and horizontal. Because the angle is 45, both of these components are the same, you can draw out the vector and see. It's a simple 45 degree right angle triangle. If the hypoteneuse is 1, then the sides are all 1/sqrt2. So if the hypoteneuse is your velocity (V), then the sides are V/sqrt2.

So we know that the ball is travelling V/sqrt2 m/s in the horizontal axis. We will call the horizontal speed Vx. It has to go 10m. distance=speed*time

10m = Vx*time

we'll isolate time, because we don't actually care what time is.

time = 10/Vx
time=10*sqrt2/Vx
(this will come in handy once we look at gravity)

Gravity accelerates everything down at 9.8m/s^2 whenever you are near the surface of the earth. It's just one of those constants you memorize. If we exclude air resistance, then gravity is the only force. So acceleration = 9.8m/s^2

What I was trying to explain earlier, is that when we throw linear acceleration into the mix, you have a different equation for position. Now there is no constant velocity because the velocity is linearly increasing/decreasing over time. So because it's linear we don't need to bother with messy integrals, we can just take the average of the initial and final velocity and use that.

In this case, the initial velocity is the vertical component(Vy) that the ball is thrown at. The final velocity is the vertical component minus the acceleration * time. This is (Vy-9.8m/s^2*time) So in equation form

Vavg = (Vy+(Vy-9.8m/s^2*time))/2
(the 9.8 is negative because gravity is acting downwards, opposite the initial velocity)
now position, or height is equal to the average velocity * time
height = Vavg*time
height = (Vy+Vy-9.8m/s^2*time)/2*time
(reorganize that a bit)
height = Vy*time/2 + Vy*time/2 - 9.8m/s^2*time*time/2
height = Vy*time - 4.9m/s^2*time^2

If the player is 2m tall and the net is 3m tall, then the ball has to achieve a final height of 1m (relative to where it starts).

I put in 1m for height and 10m/Vx for time
1m = Vy*10m/Vx - 4.9m/s^2*(10m/Vx)^2
(remember that Vy and Vx have the same value and that is V/sqrt2)
1m = 10m - 4.9m/s^2*(10m*sqrt2/V)^2
1m = 10m - 4.9*100m^2*2/V^2
9m = 4.9m/s^2*200m^2/V^2
9m*V^2=980m^3/s^2
V^2=980/9m^2/s^2
V=sqrt(980/9)m/s

V=approx 10.4m/s

This is all based on the assumption that the ball is launched from a height of 2m and the net is 3m high

It takes longer to write everything out when you try to explain it really simply. The actual equations take about a minute to do. I study mechanical
engineering, but this is all from highschool.
Click to expand...


thats beautiful you get a gold star :D
physics is one of those subjects to me imho you have to see the bigger picture then weed it out
 
Nat Turner said:
No, rockets travel in three dimensional space, encounter wind resistance, have limits on fuel capacity, etc.
Click to expand...
I'm going to back away now









:p
 
Dan said:
I'm going to assume that this is a standard 3m high basketball net and that the angle and velocity is measured from the ball leaving the player at his head height (2.00m) Now break it into vertical and horizontal.
Click to expand...
nofx says the basketball hoop is 3.05m high, though you are right that 3m is regulation.
 
Javert said:
nofx says the basketball hoop is 3.05m high, though you are right that 3m is regulation.
Click to expand...

Thanks but its cool i just subbed in 1.05 into the formulas he used.
 
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