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STD 11 - 3.2 , motion in plane — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 3.2 , motion in plane4 Marks
MCQ
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
    $\frac {u_0^2}{2g}$
  • ✓
    $\frac {2u_0v_0}{g}$
  • C
    $\frac {v_0^2}{2g}$
  • D
    $\frac {4u_0v_0}{g}$
✓

Answer

Correct option: B.
$\frac {2u_0v_0}{g}$
b
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,

$R =\frac{2 u _0 v _0}{ g }$

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