A) \[{{U}_{vertical}}\]
B) \[{{U}_{horizontal}}\]
C) \[U=\mathsf{U}_{vertical}^{2}+\mathsf{U}_{horizontal}^{2}\]
D) \[U={{(\mathsf{U}_{vertical}^{2}+\mathsf{U}_{horizontal}^{2})}^{{}^{1}/{}_{2}}}\]
Correct Answer: A
Solution :
Let a body be projected at an initial velocity\[u\] in a direction making an angle\[\theta \]with the horizontal and let it take time t to reach the highest point P of its path. The vertical velocity of the body at P is zero. From equation of motion. \[v=u-gt\] Putting, \[v={{v}_{y}}=0\]and \[u={{u}_{y}}=usin\theta \] \[0=u\text{ }\sin \theta -gt\] \[\Rightarrow \] \[t=\frac{u\sin \theta }{g}\] Hence, time of flight (T) is 2t. \[\therefore \] \[T=2t=\frac{2u\sin \theta }{g}\] \[=\frac{2}{8}\times {{U}_{vertical}}\]You need to login to perform this action.
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