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```
a ball of mass m kg is thrown vertically upwards. Another ball of mass 2m is thrown at angle θ with the vertical. Both stays in air for the same period . the heights reached by them are in the ratio of (a) 2:1 (b) 1:! © 1:cos θ (d) 1:sinθ

```
2 years ago

Arun
24741 Points
```							height obtained by an object thrown up with u velocity (h) = u^2/2g Hence, ratio of the heights (h1/h2) = u1^2/u2^2Also, since they remain in the air for same duration. So, using first law of motion v = u + atv = 0, t = -u/a, Now here, since the a is same as equal to g and t is same, so u needs to be same.
```
2 years ago
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