A projectile is launched with an initial velocity of 1.414 m/s at 45° above the horizontal. What is the vertical component of the projectile's velocity when it strikes a wall 0.50 m away?

As the projectile starts it’s motion from a vertical height at y=10m, it’s vertical motion may be described as under-

y= 10+ut-0.5gt^2

Where “u” is initial vertical velocity

= 45 sin60,

“g” is acceleration due to the gravity (10m/sec^2)

and “t” is time.

When the projectile touches the ground the vertical height “y”=0

Therefore the time “t” taken for the flight of the projectile till it touches the ground may be calculated from the above equation when y=0 as under -

0= 10+45sin60 x t- 0.5x10 t^2

0= 10+45x (√3/2)x t-5 t^2

t^2-9x (√3/2)x t-2 =0

t= [4.5√3 + √(243/4+4)]/2

t=[7.794+ 8.047]/2

t= 7.92 sec.

The horizontal distance travelled during this time=

vt

Where “v” is horizontal velocity = 45 cos60.

Therefore,

vt =45cos60 x 7.92

= 45x0.5x7.92

= 178.2 m

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle