Physics shenanigans (1 Viewer)

[VenomP]

Dont know where the formula is in my textbook. I dont recall ever being told how one might apply it to such a situation...

The velocity of a projectile 2seconds after its launch can be found from the vector diagram (on the right). What was the launch velocity of the projectile?

Here's the diagram (in words)

A right angled triangle
Horizontal side is labelled: Vh=39.2m/s
Vertical side is labelled: Vv=19.6m/s

A) 39.2m/s at 45 degrees to the horizontal
B) 43.8m/s at 30 degrees to the horizontal
C) 55.4m/s at 45 degrees to the horizontal
D) 55.4m/s at 30 degrees to the horizontal

Thanks in advance.

 

[helper]

Use pythagoras and trig.

 

[danielmarr0]

Use pythagoras and trig.

I think its 2 seconds into flight.
So the components of x and y are after 2 seconds.

Ux= Vx (Horizontal is always constant)

So

DeltaY = Ux.t+1/2.a.t^2

Find DeltaY:
DeltaY= 39.2 x2 + 1/2x -9.8x 2^2
DeltaY=98

So we Got DeltaY and can find the initial Vertical Velocity (Uy)
Because
Vy^2=Uy^2+2.a.DeltaY

So

19.6 ^2 = Uy^2 + 2x-9.8 x 98
Expanding...
384.16= Uy^2 -1920.8
Therefore
Uy^2= 384.16 + 1920.8
Uy^2= 2304.96
Uy= 48.00999896

Therefore you can say
Uy= 48
Ux= 39.2

Using Trig

48^2+39.2^2= r^2
r= 61.973ms
At Tan#= 48/39.2
#= 50degrees 45 minutes.

So why am i completely wrong....

 

[helper]

I think its 2 seconds into flight.
So the components of x and y are after 2 seconds.

Ux= Vx (Horizontal is always constant)

So

DeltaY = Ux.t+1/2.a.t^2

Should be

DeltaY = Uy.t+1/2.a.t^2

v_y = u_y + a_yt
19.6 = u_y -9.8*2
19.6 = u_y -19.6

u_y=39.2
u²=u_x² + u_y²
u²=39.2_x² + 39.2_y²
u=55.4 ms^-1