A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s”, find the maximum height
reached by the stone. What is the net displacement and the total distance covered by the stone? How long will it
take to reach the ground?.
Answers
Answer:
Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.
Acceleration due to gravity, g = 10m/s
2
(downward motion).
Maximum height, s = H.
As the body is thrown upward a = -g the relation v
2
=u
2
−2as gives v
2
=u
2
−2aH, we have,
H =
2g
u
2
−v
2
=
2(10m/s
2
)
40m/s
2
−0
2
=
20
1600
=80m.
If a stone is thrown vertically upward, it returns to its initial position after achieving maximum height.
so, the net displacement = Difference of positions between initial and final positions = 0.
Total distance covered = 80 m + 80 m = 160 m.
Hence, the displacement is 0 and the total distance covered is 160 m.
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Solved Examples
Answer:
The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.