Projectile Motion Question (1 Viewer)

[Shivorken]

this is probably not that hard, but, what would be the velocity of a ball that has traveled one metre with an angle of elevation of 10 degrees.

 

[alcalder]

Actually, easier than you think. If the incline plane has no friction (and in the perfect classroom physics world, it usually doesn't) then all you have to consider is the change in energy from gravitational potential energy to kinetic energy.

GPE = mgh

On this inclined plane, the height is given by
h = 1. sin 10^o

m = mass of ball (do you know this? It doesn't matter.)
g = 9.8 ms^-2

Then GPE at top = KE at bottom

KE = 1/2 mv²

SO

mgh = 1/2 mv²

gh = 1/2 v²

v = sqrt(2gh)

If there is friction, that is something else to consider.

Does that help?

 

[Shivorken]

very much so =D

Thank you very much alcalder

 

[Forbidden.]

OMG, all you needed was simple trig w/ right angles and a couple of formulas ??

 

[alcalder]

Yep. It was not a projectile motion question, it was a change of energy - Conservation of Energy - question. The trick to physics questions is knowing what sort of question it is

 

[Forbidden.]

Yep. It was not a projectile motion question, it was a change of energy - Conservation of Energy - question. The trick to physics questions is knowing what sort of question it is

Not those Energy Conservation problems with E_T = E_K + E_p = 0 ... but good thing I know how to use it to derive escape velocity.

 

[jyu]

Actually, easier than you think. If the incline plane has no friction (and in the perfect classroom physics world, it usually doesn't) then all you have to consider is the change in energy from gravitational potential energy to kinetic energy.

GPE = mgh

On this inclined plane, the height is given by
h = 1. sin 10^o

m = mass of ball (do you know this? It doesn't matter.)
g = 9.8 ms^-2

Then GPE at top = KE at bottom

KE = 1/2 mv²

SO

mgh = 1/2 mv²

gh = 1/2 v²

v = sqrt(2gh)

If there is friction, that is something else to consider.

Does that help?

Actually, it is even easier than you think.

It can be considered as motion with constant acceleration.

Acceleration down the inclined plane a = gsin10, u = 0, s = 1,

v^2 = u^2 + 2as = 2gsin10,

v = sqrt(2gsin10)

:wave:

 

[ads89]

Actually, easier than you think. If the incline plane has no friction (and in the perfect classroom physics world, it usually doesn't) then all you have to consider is the change in energy from gravitational potential energy to kinetic energy.

GPE = mgh

On this inclined plane, the height is given by
h = 1. sin 10^o

m = mass of ball (do you know this? It doesn't matter.)
g = 9.8 ms^-2

Then GPE at top = KE at bottom

KE = 1/2 mv²

SO

mgh = 1/2 mv²

gh = 1/2 v²

v = sqrt(2gh)

If there is friction, that is something else to consider.

Does that help?

excuse me, but is this within the scope of HSC physics? I've never seen them ask a question like this in the HSC