Answer:
In angular projection of a
projectile, let h be the height attained by projectile in timings \[{{t}_{1}}\]
and \[{{t}_{1}}\] respectively.
Then, \[h=u\sin
\text{ }\!\!\theta\!\!\text{ }\,{{t}_{1}}-\frac{1}{2}g\,t_{1}^{2}=u\sin \text{
}\!\!\theta\!\!\text{ }{{t}_{2}}-\frac{1}{2}gt_{2}^{2}\]
or\[u\sin \text{
}\!\!\theta\!\!\text{ }\left( {{t}_{2}}-{{t}_{1}} \right)=\frac{1}{2}g\left(
t_{2}^{2}-t_{1}^{2} \right)=\frac{1}{2}g\left( {{t}_{2}}-{{t}_{1}}
\right)\left( {{t}_{2}}+{{t}_{1}} \right)\]or\[\left( {{t}_{2}}+{{t}_{1}}
\right)=\frac{2u\,\sin \text{ }\!\!\theta\!\!\text{ }}{g}\]= total time of
flight.
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