The movement of a projectile, thrown upwards vertically, is described by the equation y = - 40x + 200x . Where y is the height, in meters, reached by the projectile x seconds after launch. Therefore, determine the instant when the projectile hits the ground.

sorry can't

step-by-step explanation:

i googled though and something came up but it was in science langueage not common english sooo

answer: e 9, 5.75

step-by-step explanation:

step 1

find the variation ab in the x coordinate

delta x=(x2-x1)=10-> 8 units

find the variation ab  in the y coordinate 

delta y=(y2-y1)=6-> 2 units

step 2

find the coordinates point c

  x coordinate 

cx=ax+delta x*2/> cx=2+8*(2/> cx=3

  y coordinate 

cy=ay+delta y*2/> cy=4+2*(2/> cy=4.25

the coordinates of point c are (3,4.25)

step 3

find the coordinates point d

  x  coordinate 

dx=ax+delta x*4/> dx=2+8*(4/> dx=4

  y  coordinate 

dy=ay+delta y*4/> dy=4+2*(4/> dy=4.5

the coordinates of point d are (4, 4.5)

step 4

find the coordinates point e

  x  coordinate 

ex=ax+delta x*4/> ex=2+8*(6/> ex=5

  y  coordinate 

ey=ay+delta y*4/> ey=4+2*(6/> ey=4.75

the coordinates of point e are (5, 4.75)

step 5

find the coordinates point f

  x  coordinate 

fx=ax+delta x*4/> fx=2+8*(8/> fx=6

  y  coordinate 

fy=ay+delta y*4/> fy=4+2*(8/> fy=5

the coordinates of point f are (6,5)

step 6

find the coordinates point g

  x  coordinate 

gx=ax+delta x*4/> gx=2+8*(10/> gx=7

  y  coordinate 

gy=ay+delta y*4/> gy=4+2*(10/> gy=5.25

the coordinates of point g are (7,5.25)

step 7

find the coordinates point h

  x  coordinate 

hx=ax+delta x*4/> hx=2+8*(12/> hx=8

  y  coordinate 

hy=ay+delta y*4/> hy=4+2*(12/> hy=5.5

the coordinates of point h are (8,5.5)

step 8

find the coordinates point i

  x  coordinate 

ix=ax+delta x*4/> ix=2+8*(14/> ix=9

  y  coordinate 

iy=ay+delta y*4/> iy=4+2*(14/> iy=5.75

the coordinates of point i are (9,5.75)

- r3ktforgood ☕