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Question

A ball is projected at an angle of $$ 30^0 $$ above with the horizontal from the top of a tower and strikes the ground in $$5$$ sec at an angle of $$ 45^0 $$ with the horizontal. Find the height of the tower and the speed with which it was projected.


A
Height =33.5m; Speed =36.6m/s
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B
Height =43.6m; Speed =15m/s
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C
Height =59.2m; Speed =30m/s
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D
Height =71.9m; Speed =50m/s
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Solution

The correct option is B Height $$= 33.5 m$$; Speed $$= 36.6 m/s$$
$$Let\quad initial\quad velocity\quad along\quad x\quad direction\quad be\quad { u }_{ x }\quad and\quad that\quad along\quad y\quad direction\quad be\quad { u }_{ y }\\ \dfrac { { u }_{ x } }{ cos{ 30 }^{ 0 } } =\dfrac { { u }_{ y } }{ sin{ 30 }^{ 0 } } ....(1)\\ in\quad 5\quad seconds\quad the\quad velocity\quad makes\quad an\quad angle\quad of\quad { 45 }^{ 0 }\quad with\quad the\quad horizontal,\quad which\quad means\\ horizntal\quad and\quad vertical\quad velocities\quad are\quad same.\\ { v }_{ y }={ u }_{ y }-gt=-{ u }_{ x }=-{ u }_{ y }\sqrt { 3 } \\ \Rightarrow { u }_{ y }=\dfrac { gt }{ 1+\sqrt { 3 }  } =\dfrac { 50 }{ 2.732 } =18.3\quad m/s\\ { u }_{ x }={ u }_{ y }\sqrt { 3 } =31.7\quad m/s\\ U=\sqrt { { u }_{ x }^{ 2 }+{ u }_{ y }^{ 2 } } =36.6\quad m/s$$

Physics

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