a ball of mass m is thrown vertically up another ball of mass 2m is thrown at an angle theta with the vertical . both of them stay in air for the same period of time . what is the ratio of height attained by the two balls???
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Answer:
Let h
1
be the maximum height reached by ball 1 and h
2
be the height reached by ball 2
v
1
2
=u
1
2
−2gh
1
(v
1
=0)
h
1
=
2g
u
1
2
v
2
2
=u
2
2
sin
2
θ−2gh
2
h
2
=
2g
u
2
2
sin
2
θ
Time of accent for both the bodies is same. Therefore
v
1
=u
1
−gt
1
t
1
=
g
u
1
t
2
=
g
u
2
sinθ
t
1
=t
2
(given in question)
u
1
=u
2
sinθ
h
2
h
1
=
2g
u
2
2
sin
2
θ
2g
u
1
2
Substituting u
1
=u
2
sinθ
We get h
1
=h
2
Thus, the heights attained by the two are in the ratio 1:1.
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