A) \[h/2\]
B) h
C) \[3h/2\]
D) 2h
Correct Answer: D
Solution :
In projectile motion, height \[h=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] For maximum h, \[\theta ={{90}^{o}}\] Maximum height \[=\frac{{{u}^{2}}}{2g}=h\] Maximum range is obtained when angle of projection is \[{{45}^{o}}\]. Maximum range \[=\frac{{{u}^{2}}}{g}=2h\]You need to login to perform this action.
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