#1
So I'm having some issues solving this physics problem dealing with motion in multiple dimensions.

A 2.00-m-tall basketball player wants to make a goal from 10.0 m from the basket. If he shoots the ball at a 45 degree angle, at what initial speed must he throw the basketball so that it goes thought he hoop without striking the backboard. The hoop is 3.05 m high.


I haven't done much complicated physics yet. All I've done so far are the basic equations of motion: v = v0 + a t, x = v0 t + 1/2 a t^2, v^2 = v0^2 + 2 a x, that kind of stuff. In terms of trigonometry, I don't use anything but the basic sin, cos, tan stuff.
Last edited by LiquidTension99 at Sep 29, 2008,
#2
and thats all you need, seeing as the ball should travel in a perfect arc (in theory). and you already know the slope at one point, and the predicted slope at the other, so it shouldn't be that hard.
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#3
Quote by muse-ik
and thats all you need, seeing as the ball should travel in a perfect arc (in theory). and you already know the slope at one point, and the predicted slope at the other, so it shouldn't be that hard.


Well I believe you when you say that I have all that I need, but I've never talked about the slope of the shot or whatever. Just stuff relating to speed, acceleration, time, and distance. Maybe that's what you mean, but I need some clarification. I know it's not super-advanced, I sort of just don't know where to start on this one.
#4
ill get my old yr 12 physics book and get back to you
Sig.



After "bio class" did you and your classmates organize a "Celebrity Skateboard Tournament" to save your beloved "Rec Center" where "Coach Anderson" taught you how to become men?
#5
a= acceleration
u= initial velocity
s= displacement
t= time

displacement in x direction = 10
displacement in y direction = 3.05-2 = 1.05

initial velocity in x direction = u cos45
initial velocity in y direction = u sin45 = u cos(90-45) = u cos 45

using s = ut + 0.5at^2,
in x direction: 10 = (u cos45)(t) + 0.5(0)(t^2) ...Eqn 1
in y direction: 1.05 = (u cos45)(t) + 0.5(-9.81)(t^2) ...Eqn 2

Eqn 1 - Eqn 2,
10-1.05 = -(-4.905)(t^2)
t = 1.351

sub t = 1.351 into Eqn 1,
u cos45 = 7.402
u = 10.468

I think its right.. would someone double check?
#6
i believe that is correct ^^^

edit: nice work btw dude
Sig.



After "bio class" did you and your classmates organize a "Celebrity Skateboard Tournament" to save your beloved "Rec Center" where "Coach Anderson" taught you how to become men?
Last edited by boj at Sep 30, 2008,
#7
Quote by xyber56
a= acceleration
u= initial velocity
s= displacement
t= time

displacement in x direction = 10
displacement in y direction = 3.05-2 = 1.05

initial velocity in x direction = u cos45
initial velocity in y direction = u sin45 = u cos(90-45) = u cos 45

using s = ut + 0.5at^2,
in x direction: 10 = (u cos45)(t) + 0.5(0)(t^2) ...Eqn 1
in y direction: 1.05 = (u cos45)(t) + 0.5(-9.81)(t^2) ...Eqn 2

Eqn 1 - Eqn 2,
10-1.05 = -(-4.905)(t^2)
t = 1.351

sub t = 1.351 into Eqn 1,
u cos45 = 7.402
u = 10.468

I think its right.. would someone double check?
beat me to it
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#8
Quote by boj
i believe that is correct ^^^

edit: nice work btw dude

Haha thanks man