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?
Answers
Answer:
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 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.
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 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.