A ball initially at rest is dropped from a window sill 14.2 m off the ground. Find the velocity of the ball the moment it hits the ground. Round your answer to 2 decimals places.

16.68 m/s

Explanation:

The following data were obtained from the question:

Height (h) = 14.2 m

Initial Velocity (u) = 0 m/s

Final velocity (v) =?

Thus, the velocity (v) with which the ball hit the ground can be obtained as follow:

v² = u² + 2gh

NOTE: the acceleration due to gravity (g) = 9.8 m/s²

v² = u² + 2gh

v² = 0² + 2 × 9.8 × 14.2

v² = 0 + 278.32

v² = 278.32

Take the square root of both side

v = √278.32

v = 16.68 m/s

Therefore, the velocity with which the ball hit the ground is 16.68 m/s

i have no clue hope you find the answer

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole