A stone is thrown vertically upwards with a velocity 30 ms. If the acceleration due to gravity is 10 mas - what is the distance travelled by the particle during the first second of its motion?Find its maximum height?
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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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