- #1
Lisa...
- 189
- 0
Hey!
I have the following problem to solve:
Consider a ball that is thrown with initial speed v0 at an angle a above the horizontal. If we consider it's speed at some height h above the ground show that v(h) is independent of a.
I started to solve the problem this way:
v= sqrt (vx^2 + vy^2)
vx= v0x= v0 cos a
vy= v0y- gt= v0 sin (a) -gt
Substitution of vx and vy in the first formula gives:
v= sqrt(v0^2 (cos^2)a + v0^2 ((sin^2)a) - 2 v0 gt sin a + g^2 t^2)
v^2= v0^2 (cos^2)a + v0^2 ((sin^2)a) - 2 v0 gt sin a + g^2 t^2
v^2= v0^2 ((cos^2)a + (sin^2)a) -2 v0 gt sin a + g^2 t^2
v^2= v0^2 -2 v0 gt sin a + g^2 t^2
Now I STILL have a term with a in it (-2 v0 gt sin a ) How in the world will I get rid of it?!
I have the following problem to solve:
Consider a ball that is thrown with initial speed v0 at an angle a above the horizontal. If we consider it's speed at some height h above the ground show that v(h) is independent of a.
I started to solve the problem this way:
v= sqrt (vx^2 + vy^2)
vx= v0x= v0 cos a
vy= v0y- gt= v0 sin (a) -gt
Substitution of vx and vy in the first formula gives:
v= sqrt(v0^2 (cos^2)a + v0^2 ((sin^2)a) - 2 v0 gt sin a + g^2 t^2)
v^2= v0^2 (cos^2)a + v0^2 ((sin^2)a) - 2 v0 gt sin a + g^2 t^2
v^2= v0^2 ((cos^2)a + (sin^2)a) -2 v0 gt sin a + g^2 t^2
v^2= v0^2 -2 v0 gt sin a + g^2 t^2
Now I STILL have a term with a in it (-2 v0 gt sin a ) How in the world will I get rid of it?!