A stone is throw vertically upward direction and reaches the ground in 16 seconds. Calculate the maximum height it reached. Take acceleration due to gravity=10m/s.
Answers
Explanation:
Initial Velocity u=40
Fianl velocity v=0
Height, s=?
By third equation of motion
v
2
−u
2
=2gs
0−40
2
=−2×10×s
s=
20
160
⇒s=80m/s
Toatl distance travelled by stone = upward distance + downwars distance =2×s=160m
Total Diaplacement =0, Since the initial and final point is same
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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