An object is thrown into the air with an initial velocity of 12 meters per seconds from a platform that is raised 4 meters above the ground. how long will it take for the object to return to the ground?

1.52 seconds

Explanation:

Step 1: identity the given parameters

Initial velocity (u) = 12m/s

Height above ground (h1) = 4m

Final velocity (V) = 0

Step 2: calculate the height travelled by the object from 4m height (h2).

V^2 = U^2 -2gh

0= 12^2-2(9.8*h)

2(9.8*h) = 12^2

19.6*h = 144

h = 144/19.6

h = 7.347 m

Total height above ground (ht) = 4m +7.347m = 11.347m

Step 3: calculate the time reach ground

T = √(2h/g)

T = √(2*11.347/9.8)

T= √(22.694/9.8)

T= √2.316

T= 1.52 seconds

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.