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Answer :
Given, initial velocity = 5 m/s
Final velocity = 0 m/s
{since the stone will stop at maximum height and then again starts falling down}
\( a = 10 \ m/s^2 \) in the direction opposite to the trajectory of the stone \( = -10 \ m/s^2 \)
By the third motion equation,
\(\therefore v^2 - u^2 = 2as \)
\( \Rightarrow s = 1.25 m \)
\(\therefore \) The distance travelled by the stone = 1.25 meters
Now, using the first motion equation, we get
\( v = u + at \) \( t = 0.5 s \)
Therefore, time taken by the stone to reach a maximum height = 0.5 seconds