A 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?

Given that

  • 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

Find out

The height attained by the stone and how much time will it take to reach there

Formula

As per the third motion equation,

v2 – u2 = 2as

Therefore, the distance travelled by the stone is

(s) = (02 – 52)/ 2(10)

Distance (s) = 1.25 meters

As per the first motion equation,

v = u + at

Therefore, time is 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.

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