5. A cricket ball is thrown up. It reaches a height of
10 m from the point of throw and then reaches
back to the point of throw. Find the distance and
displacement of the ball.
Answers
Answer:
Given, s=20m,v=0m/sec,a=9.8m/s
2
1) Initial velocity
v
2
−u
1
=2as
⇒0−u
2
=2(9.8)(20)
u
2
=393m/sec
⇒u=19.8m/sec
2) Final velocity
u=0m/sec
v=?
a=9.8m/s
2
s=20m
v=19.8m/s
3)S
1
=?,t
1
=0.5sec,S
2
=?t
2
=2.5sec
For S
1
=ut
1
+
2
1
a(t
1
)
2
=19.8×0.5+
2
1
(−9.8)(0)
2
=8.675m
For S
2
=ut
2
+
2
1
a(t
2
)
2
=19.8×2.5+
2
1
(−9.8)(2.5)
2
=18.875m
Distance travelled=18.875−8.675=10.2m
4)t=?,u=19.8,s=15m,a=−9.8m/s
2
⇒S=ut+
2
1
(a)(t
2
)
⇒4.9t
2
−19.8t+15=0
⇒t=
2×4.9
19.8±9.8
=1.01s or 3.03s
∴ When going up, t=3.03s
When coming down, t=1.01s
Answer:
Distance is 20m
Displacement is 0
Explanation:
The distance covered by ball =
From point of throw to a certain height = 10m
From a certain height to point of throw = 10m
Therefore,
10 + 10 = 20m
Displacement = 0 (Because the object didn't end in a different point . it came back to the point of throw. )
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