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 answer:
Answer:
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
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