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3-2.Motion in Plane
medium
A projectile is projected from ground with initial velocity $\vec u\, = \,{u_0}\hat i\, + \,{v_0}\hat j\,$. If acceleration due to gravity $(g)$ is along the negative $y-$ direction then find maximum displacement in $x-$ direction
A
$\frac {u_0^2}{2g}$
B
$\frac {2u_0v_0}{g}$
C
$\frac {v_0^2}{2g}$
D
$\frac {4u_0v_0}{g}$
Solution
Given,
$\overrightarrow{ U }=u_0 \hat{1}+v_0 \hat{j}$
$u_0=u \cos \theta \ldots \ldots .1$
$v_0=v \sin \theta \ldots \ldots . .2$
Since,
$R =\frac{ u ^2 2 \sin 2 \theta}{ g }$
$R =\frac{ u ^2 2(\sin \theta \cdot \cos \theta)}{ g } \ldots \ldots$
Substituting the value of equation 1,2 in eqyuation 3 then we get,
$R =\frac{2 u _0 v _0}{ g }$
Standard 11
Physics
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