Aball is thrown directly upward from a height of 7 feet with an initial velocity of 50 ft./s use the quadratic formula to find the number of seconds it takes the ball to hit the ground

The time taken by ball to hit the ground is 5.10 seconds .

Step-by-step explanation:

Given as :

The height cover by ball to move upward = 7 feet

The initial velocity = u = 50 feet per sec

Let the time taken by ball to hit the ground = t sec

Let the deacceleration rate = - 9.8 m/s²

The final velocity of the ball = v=  0 at the top

So, From equation

v = u + at

I.e v = u + a × t

Or, 0 = 50 + ( - 9.8) × t

Or, 9.8 t = 50

∴ t =

I.e t = 5.10 seconds

So, time taken = 5.10 sec

Hence The time taken by ball to hit the ground is 5.10 seconds . Answer

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1