Question

a stone is throw vertically upward with a velocity of 19.6m/s^.calculate the distance and displacement after 3sec with acceleration of 9.8 m/sec^​

Answer 1

Answer:

14.7m

Explanation:

initial velocity (u) = 19.6 m/s

time(t) = 3s

displacement(S) = ut +(-1/2 gt²)

S = (19.6×3) -(1/2×9.8×9)

S = 58.8 - 44.1 = 14.7m

distance = displacement= 14.7m

(because the stone is moving in only one direction for 3 secs)

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