The speed of a projectile at its maximum heig | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 ......... $^o$

Similar Questions

From the top of a tower of height $40\, m$, a ball is projected upwards with a speed of $20\, m/s$ at an angle $30^o$ to the horizontal. The ball will hit the ground in time ......... $\sec$ (Take $g = 10\, m/s^2$)

A ball is projected from ground with a speed of $20\,m / s$ at an angle of $45^{\circ}$ with horizontal. There is a wall of $25\,m$ height at a distance of $10\,m$ from the projection point. The ball will hit the wall at a height of $.........\,m$

The equation of a projectile is $y=\sqrt{3} x-\frac{ x^2}{2}$, the velocity of projection is

A ball of mass $m$ is thrown vertically upward. Another ball of mass $2\,m$ is thrown an angle $\theta$ with the vertical. Both the balls stay in air for the same period of time. The ratio of the heights attained by the two balls respectively is $\frac{1}{x}$. The value of $x$ is $.....$

The equation of motion of a projectile are given by $ x = 36 t\,metre$ and $2y = 96 t -9.8 t^2\, metre$. The angle of projection is