A) \[\frac{gt}{2\sin \left( \alpha -\beta \right)}\]
B) \[\frac{gt\,\cos \beta }{2\sin \left( \alpha -\beta \right)}\]
C) \[\frac{\sin \left( \alpha -\beta \right)}{2gt}\]
D) \[\frac{2\sin \left( \alpha -\beta \right)}{gt\,\cos \beta }\]
Correct Answer: B
Solution :
[b] \[t=\frac{2u\sin \left( \alpha -\beta \right)}{g\cos \beta }.\text{ };\text{ }u=\frac{gt\cos \beta }{2\sin \left( \alpha -\beta \right)}\]You need to login to perform this action.
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