The speed of a projectile at its maximum heig | vedclass

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Question

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,Let\,v\,be\,velocity\,of\,projectile\,at\
{m{maximum}}\,{m{height}}\,{m{H}}{m{.}}\
\,\,\,\,\,\,

Vedclass · Accepted Answer

$60$

Vedclass · Answer

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,Let\,v\,be\,velocity\,of\,projectile\,at\
{m{maximum}}\,{m{height}}\,{m{H}}{m{.}}\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,v\, = \,u\cos 	heta \
According\,to\,given\,problem,\,v = \frac{u}{2}\
	herefore \,\frac{u}{2} = \,u\cos \,	heta  \Rightarrow \cos 	heta \frac{1}{2} \Rightarrow 	heta  = {60^ \circ }
\end{array}$

The speed of a projectile at its maximum height is half of its intial speed. The angle of projection is ......... $^o$

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