An object is thrown vertically upwards with velocity of 49 m/s calculate the hight attained by the object & also time required to attend in the height if the acceleration is 9.8m/s square
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Explanation:
Consider a formula,
2gH = v2 – u2
2 × (- 9.8) × H = 0 – (49)2
– 19.6 H = – 2401
H = 122.5 m
Now consider a formula,
v = u + g × t
0 = 49 + (- 9.8) × t
– 49 = – 9.8t
t = 5 sec
(1) The maximum height to which the ball rises = 122.5 m
(2) The total time ball takes to return to the surface of the earth = 5 + 5 = 10 sec.
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