A ball is thrown upward with an initial velocity of 11 meters per second. Suppose that the initial height above the ground is 6 meters. At what time will the ball hit the ground? The ball is on the ground when s=0. Use the equation s=-5t^2+vt+h

i believe

step-by-step explanation:

from my conclusion c

option 1 is correct.

step-by-step explanation:

y varies inversely with x and constant of variation is 4

the equation will be:

y = k/x

where k =4 so,

y = 4/x

if we put values of x of option 1 in the above equation and find values of y

y = 4/-2 => y =-2

y = 4/-1 => y = -4

y = 4/1 => y = 4

y = 4/2 => y =2

so, all the values of x and y are correct.

if we put values of x given in other options and find value of y we cannot find whole table satisfying the equation.

hence option 1 is correct.