Question

a ball is thrown upwards with a speed of 14 metre per second how much high will it go before it starts falling​

Answer 1

Answer:

Explanation:

The correct answer is 1.43 s.

The correct answer is 1.43 s.

The correct answer is 1.43 s.

Answer 2

Answer:

HELLO!

$...$

Vf^2 = Vi^2 +/- 2aS

Vf is Velocity final which is zero at maximum height.

Vi is Velocity initial which is 14 m/s

a or g is gravitational acceleration which is a constant (9.81 m/s^2). This constant varies if your teacher has a different value for it.

S is the maximum height

+/- means (+) is falling and (-) if the object is rising

so putting the values in the formula:

0 = 14^2 - 2(9.81)(S)

0 = 14^2 - 19.62(S)

-196 = -19.62(S)

S = 9.98 m → Ans.

HOPE IT HELPS ARMY :)

PURPLE YA