Aprojectile is fired at time t = 0.0 s, from point 0 at the edge of a cliff, with initial velocity components of v ax= 50m/s and v ay= 200m/s. the projectile rises, then falls into the sea at point p. the time of flight of the projectile is 50.0s what is the y-coordinate of the projectile when its x-coordinate is 1000 m?

The projectile has constant horizontal speed all the time and we can use this formula

x=Vax*t, where t is time it (projectile) takes to reach x=1000m

t=x/Vax=1000/50=20s

As we know we can split projectile trajectory in two parts and the time for each part is equal. We can conclude that time t=20s belong to first part where projectile is going up.

Formula for this type of movement is  y=Vay-(gt∧2)/2

If we take acceleration of gravity g=10m/s∧2

we get     y=200*20-(10*20∧2)/2=4000-2000=2000m

But if I calculate time it takes the highest point I get  

t=Vay/g=200/10=20s

The highest point for that time is

y= H=Vay∧2/2g=200∧2/2*10=40000/20=2000m

This happened because the projecile is fired from the edge of a cliff which is above the sea line.

Good luck!!!

connie < felicia < stew

step-by-step explanation:

stew:

we need to find the slope

m = (y2-y1)/ (x2-x1)

    = (2650-795)/(10-3)

    = 1855/7

265 words per minute

connie:

we need to find the slope

m = (y2-y1)/ (x2-x1)

we have two points (0,0) and (4,800)

m = (800-0)/(4-0)

  m = 800/4

    200 words per minute

felicia:

260 words per minute

200< 260 < 265

connie < felicia < stew