Physics Two Dimensional Motion Question 58
Question: A stone projected with a velocity $ u $ at an angle $ \theta $ With the horizontal reaches maximum height $ H _{1} $ When it is projected with velocity u at an angle $ (\frac{\pi }{2}-\theta ) $ with the horizontal, it reaches maximum height $ H _{2} $ . The relation between the horizontal range R of the projectile, $ H _{1} $ and $ H _{2} $ is
Options:
A) $ R=4\sqrt{H _{1}H _{2}} $ B) $ R=,4(H _{1}-H _{2}) $ C) $ R=,4(H _{1}+H _{2}) $ D) $ R=,\frac{H^{2} _{1}}{H^{2} _{2}} $
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Answer:
Correct Answer: A
Solution:
[a] $ H _{1}=\frac{u^{2}{{\sin }^{2}}\theta }{2g}and,,,H _{2}=\frac{u^{2}{{\sin }^{2}}(90-\theta )}{2g}=\frac{u^{2}{{\cos }^{2}}\theta }{2g} $
$ H _{1}H _{2}=\frac{u^{2}{{\sin }^{2}}\theta }{2g}\times \frac{u^{2}{{\cos }^{2}}\theta }{2g} $
$ =\frac{{{(u^{2}sin2\theta )}^{2}}}{16g^{2}}=\frac{R^{2}}{16} $
$ [\
Therefore R=4\sqrt{H _{1}H _{2}} $
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