Question

If two balls are projected at angle 45degrees and 60 degrees and maximum height reached are same , what is rationof initial velocities

Answer 1

Solution :

Let the balls A and B

Maximum height of ball A

$\sf h = \frac{ {u}^{2} {sin}^{2} \theta}{2g} \\ \\ \sf h = \frac{ {u}^{2} { (\sin45 )}^{2}}{2g} \\ \\\sf h = \frac{ {u}^{2} }{2g} \times { \bigg( \frac{1}{2} \bigg) }^{2} \\ \\\sf h = \frac{ {u}^{2} }{4g} \\ \\\sf {u}^{2} = 4hg \\ \\ \boxed {\sf \blue{u = \sqrt{4hg} }}$

Maximum height of ball B

$\sf h = \frac{ {u}^{2} { (\sin 60)}^{2} }{2g} \\ \\\sf h = \frac{ {u}^{2} }{2g} \times { \bigg( \frac{ \sqrt{3} }{2} \bigg) }^{2} \\ \\\sf h = \frac{ {3u}^{2} }{8g} \\ \\ \sf {u}^{2} = \frac{8gh}{3} \\ \\\boxed{\sf \green{ u = \sqrt{ \frac{8gh}{3}}} }$

Ratio of the Initial velocity

$\sf \frac{ \sqrt{4gh} }{ \sqrt{ \frac{ 8gh }{3}} } \\ \\ \sf \frac{ \sqrt{4gh} \times\sqrt{ 3 }}{ \sqrt{8gh} } \\ \\ \sf \frac{ \sqrt{3} }{ \sqrt{2} }$

$\sf \orange{Ratio \: of \: initial \: velocity\: is\: \sqrt3 : \sqrt2}$