Question

A stone is projected with a velocity u at angle $ \theta $ with the horizontal reaches maximum height H1, when it is projected with a velocity u at an angle $ \left( \frac{\pi }{2}-\theta \right) $ with the horizontal, it reaches to a maximum height H2. The relation between the horizontal range R of the projectile, H1 and H2 is:

• A. $ R=4\sqrt{{{H}_{1}}{{H}_{2}}} $
• B. $ R=4({{H}_{1}}-{{H}_{2}}) $
• C. $ R=4({{H}_{1}}+{{H}_{2}}) $
• D. $ R={{H}_{1}}^{2}/{{H}_{2}}^{2} $

Answer 1

Answer: (A)

$ {{H}_{1}}=\frac{{{u}^{2}}si{{n}^{2}}\theta }{2g} $ $ {{H}_{2}}=\frac{{{u}^{2}}{{\sin }^{2}}(90-\theta )}{2g}=\frac{{{u}^{2}}{{\cos }^{2}}2\theta }{2g} $ From $ \frac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{T}_{2}}{{V}_{2}}}{{{T}_{2}}} $ $ \frac{{{P}_{1}}{{m}_{1}}}{\rho {{T}_{1}}}=\frac{{{P}_{2}}{{m}_{2}}}{\rho {{T}_{2}}} $ $ \frac{720\times m}{313}=\frac{{{P}_{2}}\left( m-\frac{1}{4}m \right)}{626} $ $ \frac{720\times m}{313}=\frac{{{P}_{2}}\times \frac{3}{4}m}{626} $ $ {{P}_{2}}=\frac{720\times 626\times 4}{313\times 3} $ $ =1920\,kPa. $