A stone is thrown vertically upward with an initial velocity of 40m/s . Taking g=10m/s , find the maximum height reached by the stone.what is the net displacement and the total distance covered by the stone?
Answers
A stone is thrown with speed 40 m/s , at highest point velocity of stone = 0 m/sec
use , kinematics equation ,
V² = U² + 2aS
where V = 0
U = 40 m/s
a = -g m/s² = -10 m/s²
0 = (40)² -2 × 10 × S
S = 80 m .
hence maximum height is reached by stone = 80 m .
now, stone comes back to initial point.
hence
total distance covered by stone = 80 ( in upward motion ) + 80( in downward motion ) = 160 m .
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hey
the answer
According to the equation of motion under gravity v2 − u2 = 2gs
Where,
u = Initial velocity of the stone = 40 m/s
v = Final velocity of the stone = 0 m/s
s = Height of the stone
g = Acceleration due to gravity = −10 ms−2
Let h be the maximum height attained by the stone.
Therefore, 0 2 − 402 = 2(−10)ℎ ⇒ ℎ = (40×40) / 20 = 80
Therefore, total distance covered by the stone during its upward and downward
journey = 80 + 80 = 160 m
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