Question

stone is thrown in a vertically upward direction with a velocity of 5 m s-1. If the acceleration of the stone during its motion is 10 m s–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?​

Answer 1

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✎Given, initial velocity (u) = 5 m/s

Terminal velocity (v) = 0 m/s (since the stone will reach a position of rest at the point of maximum height)

✎Acceleration = 10 ms-2 in the direction opposite to the trajectory of the stone = -10 ms-2

As per the third motion equation, v2 – u2 = 2as

Therefore, the distance traveled by the stone (s) = (02 – 52)/ 2(10)

✎Distance (s) = 1.25 meters

As per the first motion equation, v = u + at

Therefore, time taken by the stone to reach a position of rest (maximum height) = (v – u) /a

=(0-5)/-10 s

Time taken = 0.5 seconds

Therefore, the stone reaches a maximum height of 1.25 meters in a timeframe of 0.5 seconds.✓✓

Answer 2

Answer:

0.5

is the answer of that question