velocity question (1 Viewer)

[jonnypride]

does anyone know this question

a shot is released at an angle of 35degrees with horizontal and and has a velocity of 6m/sec along line of projection. there is no air resistance.

what is the vertical component of velocity

1) at instant release ( i know that its 0)
2) at the highest point:?
3) after 0.05 second

 

[Trebla]

does anyone know this question

a shot is released at an angle of 35degrees with horizontal and and has a velocity of 6m/sec along line of projection. there is no air resistance.

what is the vertical component of velocity

1) at instant release ( i know that its 0)
2) at the highest point:?
3) after 0.05 second

Is this supposed to be a projectile motion question? (It's not in the 2 unit course)
The vertical component of the velocity is NOT zero at the instant it is released. It is zero at its highest point, because it changes sign from positive to negative at that point and hence must pass through zero. You work out the vertical component of velocity at the instant of release and after 0.05 seconds by deriving the equations of motion for y (= Vtcosθ - gt²/2).

 

[littlebigpsycho]

physics equations are applicable but im too sick of physics atm so cbf

dunno why its in your maths course unless its x1/2

 

[jonnypride]

Is this supposed to be a projectile motion question? (It's not in the 2 unit course)
The vertical component of the velocity is NOT zero at the instant it is released. It is zero at its highest point, because it changes sign from positive to negative at that point and hence must pass through zero. You work out the vertical component of velocity at the instant of release and after 0.05 seconds by deriving the equations of motion for y (= Vtcosθ - gt²/2).

hi thanks for your help.

i was just wondering so if im trying to gind tghe vertical component of velocity after 0.05 seconds using the formula u gave me

Does the T stand for the time (0.05) or the time the projectile spent in the air?

thanks