Aball is thrown straight up with a speed of 12 meters per second near the surface of earth. what is the maximum height reached by the ball?

maximum height =7.347m

Step-by-step explanation:

maximum height = (U²sin²θ)/2g

where θ = 90° as the ball is thrown straight up.

sin90°=1 , so our formula reduces to;

H= U²/2g

U=12m/s  , g=9.8m/s²

H= 12²/(2*9.8)

H=7.347m

H=7.339m

Step-by-step explanation:

The formula to find maximum height is :

H= v^2/(2× g)

V means speed=12 m/s

g means acceleration due gravity (constant)= 9.81m/s^2

Apply the formula

H=(12^2)/(2×9.81)

H=144/19.62

H=7.339m

i dont know

step-by-step explanation:

answer: integers are just whole numbers: positive, negative, and zero.

i = {-∞, -2, -1, 0, 1, 2, 3, ∞}

let x = 1st integer

let y = 2nd integer

x = 3y + 17   {equation 1}

xy = -24         {equation 2}

substitute 3y+17 in place of x n equation 2

and solve for y.

(3y+17)y = -24

3y2 + 17y = -24

3y2 + 17y + 24 = 0

from quadratic equation:

y = [-17±√(172-4(3)(24))]/(2(3))

y = (-17 ±√1)/6

y = (-17+1)/6   or     y = (-17-1)/6

y = -16/6           or     y = -18/6

y = -2 2/3         or   y = -3

since we are dealing with

y = -3

x = 3y+17 = -9+17 = 8

the integers are -3, 8

step-by-step explanation: