A) \[40{}^\circ C\]
B) \[90{}^\circ C\]
C) \[30{}^\circ C\]
D) \[45{}^\circ C\]
Correct Answer: C
Solution :
Suppose \[I\omega \] is the angle of projection and u is the initial velocity, then Horizontal range \[4\sqrt{3}\] or \[\frac{T}{\sqrt{3}}\] Maximum height is, \[\frac{T}{3}\] Given \[\frac{\sqrt{3}}{2}T\] \[\sqrt{3}T\] \[1.6\times {{10}^{-19}}\] or \[1.12\times {{10}^{-19}}\] So, \[1.76\times {{10}^{-19}}\]
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