A) 5 : 4
B) 5 : 2
C) 5 : 1
D) 10 : 1
Correct Answer: A
Solution :
Height of projectile \[{{H}_{\max }}=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] Time of flight \[T=\frac{2u\sin \theta }{g}\] \[\therefore \] \[\frac{{{H}_{\max }}}{{{T}^{2}}}=\frac{{{u}^{2}}{{\sin }^{2}}\theta /2g}{{{(2u\sin \theta /g)}^{2}}}\] \[=\frac{g}{g}=\frac{10}{8}\] \[\therefore \] \[{{H}_{\max }}:{{T}^{2}}=5:4\]
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