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,
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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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.
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?
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?
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,
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