A) twice the vertical height
B) vertical height
C) four times the vertical height
D) three times the vertical height
Correct Answer: C
Solution :
Let \[u\]be the velocity of projection and 6 is angle of projection, then \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] and \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{g}\] Given \[\theta ={{45}^{o}}\] \[\therefore \] \[R=\frac{{{u}^{2}}\sin {{90}^{o}}}{g}=\frac{{{u}^{2}}}{g}\] ?(i) \[H=\frac{{{u}^{2}}{{\sin }^{2}}{{45}^{o}}}{g}=\frac{{{u}^{2}}}{4g}\] From Eqs. (i) and (ii), we getYou need to login to perform this action.
You will be redirected in
3 sec