18. A ball thrown up vertically returns to the thrower after 6 s.
Find
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after 4 s.
Answers
Explanation:
11th
(a) The velocity with which it was thrown up,
(b) The maximum height it reaches, and
(c) Its position after 4s.
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ANSWER
The ball returns to the ground after 6 seconds.
Thus the time taken by the ball to reach to the maximum height (h) is 3 seconds i.e t=3 s
Let the velocity with which it is thrown up be u
(a). For upward motion,
v=u+at
∴ 0=u+(−10)×3
⟹u=30 m/s
(b). The maximum height reached by the ball
h=ut+
2
1
at
2
h=30×3+
2
1
(−10)×3
2
h=45 m
(c). After 3 second, it starts to fall down.
Let the distance by which it fall in 1 s be d
d=0+
2
1
at
′2
where t
′
=1 s
d=
2
1
×10×(1)
2
=5 m
∴ Its height above the ground, h
′
=45−5=40 m
Hence after 4 s, the ball is at a height of 40 m above the ground.
Answer:
Here is your answer...
Explanation:
Answer
The ball returns to the ground after 6 seconds.
Thus the time taken by the ball to reach to the maximum height (h) is 3 seconds i.e t=3 s
Let the velocity with which it is thrown up be u
(a). For upward motion,
v=u+at
∴ 0=u+(−10)×3
⟹u=30 m/s
(b). The maximum height reached by the ball
h=ut+
2
1
at
2
h=30×3+
2
1
(−10)×3
2
h=45 m
(c). After 3 second, it starts to fall down.
Let the distance by which it fall in 1 s be d
d=0+
2
1
at
′2
where t
′
=1 s
d=
2
1
×10×(1)
2
=5 m
∴ Its height above the ground, h
′
=45−5=40 m
Hence after 4 s, the ball is at a height of 40 m above the ground.
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