A ball projected from ground at an angle of 4 | vedclass

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

Question

$\begin{array}{l}
As\,ball\,is\,projected\,at\,an\,angle\,{45^{ \circ \,}}to\,the\
horizontal\,therefore\,Rang = 4H\
or\,\,\,\,10 = 4H \Rightarro

Vedclass · Accepted Answer

$2.4$

Vedclass · Answer

$\begin{array}{l}
As\,ball\,is\,projected\,at\,an\,angle\,{45^{ \circ \,}}to\,the\
horizontal\,therefore\,Rang = 4H\
or\,\,\,\,10 = 4H \Rightarrow H\frac{{10}}{4} = 2.5m\
\left( {Range\, = 4m\, + 6\,m = 10m} ight)\
Maximum\,height,\,H = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}}\
	herefore \,{u^2} = \frac{{H 	imes 2g}}{{{{\sin }^2}	heta }} = \frac{{2.5 	imes 2 	imes 10}}{{{{\left( {\frac{1}{{\sqrt 2 }}} ight)}^2}}} = 100
\end{array}$

$\begin{array}{l}
or,\,u = \sqrt {100}  = 10\,m{s^{ - 1}}\
Height\,of\,wall\,PA\
 = OA\,	an \,	heta  - \frac{{1\,g{{\left( {OA} ight)}^2}}}{{2\,{u^2}\,{{\cos }^2}	heta }}\
 = 4 = \frac{1}{2} 	imes \frac{{10 	imes 16}}{{10 	imes 10 	imes \frac{1}{{\sqrt 2 }} 	imes \frac{1}{{\sqrt 2 }}}} = 2.4m
\end{array}$

A ball projected from ground at an angle of $45^o$ just clears a wall in front. If point of projection is $4\,m$ from the foot of wall and ball strikes the ground at a distance of $6\,m$ on the other side of the wall, the height of the walI is ........ $ m$

Similar Questions

The initial velocity of a particle of mass $2\,kg$ is $(4 \hat{ i }+4 \hat{ j })\,m / s$. A constant force of $-20 \hat{ j }\,N$ is applied on the particle. Initially, the particle was at $(0,0)$. Find the $x$-coordinate of the point where its $y$-coordinate is again zero.$..........\,m$

At the highest point of the path of a projectile, its

Figure shows four paths for a kicked football. Ignoring the effects of air on the flight, rank the paths according to initial horizontal velocity component, highest first

The velocity of projection of a body is increased by $2 \% .$ Other factors remaining unchanged, what will be the percentage change in the maximum height attained ? (in $\%$)

A ground to ground projectile is at point $A$ at $t=\frac{T}{3}$, is at point $B$ at $t=\frac{5 T}{6}$ and reaches the ground at $t=T$. The difference in heights between points $A$ and $B$ is