A) \[\frac{2{{u}^{2}}h}{g{{b}^{2}}}\]
B) \[\frac{{{u}^{2}}h}{g{{b}^{2}}}\]
C) \[\frac{2{{u}^{2}}b}{gh}\]
D) \[\frac{{{u}^{2}}b}{gh}\]
Correct Answer: A
Solution :
Coordinate of point \[(nb,nh)\] and \[\theta =0\] From equation of projectile, \[y=x\tan \theta -\frac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }\] \[nh=nb\tan 0{}^\circ -\frac{g{{n}^{2}}{{b}^{2}}}{2{{u}^{2}}{{\cos }^{2}}0{}^\circ }\] \[nh=\frac{g{{n}^{2}}{{b}^{2}}}{2{{u}^{2}}}\] \[n=\frac{2h{{u}^{2}}}{g{{b}^{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
