Projectile for PROs

\(Find\quad the\quad value\quad of\quad \theta \quad for\quad which\quad \quad the\quad area\quad b/w\quad \\ trajectary(\quad path\quad of\quad projectile)\quad and\quad the\quad ground\quad is\quad maximum\quad .\\ Eqations\quad are\quad given-:\\ Time\quad of\quad flight\quad =\quad 2u\sin { \theta } /g\\ Maximum\quad Height\quad =\quad { u }^{ 2 }{ (\sin { \theta } ) }^{ 2 }/2g\\ Range\quad =\quad 2{ u }^{ 2 }\sin { \theta } cos{ \theta }/g\\ Eqation\quad of\quad trajectory:\quad y\quad =\quad x\tan { \theta } -\frac { g{ x }^{ 2 } }{ 2{ u }^{ 2 }{ (cos{ \theta }) }^{ 2 } } \\ answer\quad in\quad degrees.\quad \)

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