A body of mass $10 kg$ is projected at an angle of $45^{\circ}$ with the horizontal. The trajectory of the body is observed to pass through a point $(20,10)$. If $T$ is the time of flight, then its momentum vector, at time $t =\frac{ T }{\sqrt{2}}$, is.
$\left[\right.$ Take $\left.g=10 m / s ^{2}\right]$
A body of mass $10 kg$ is projected at an angle of $45^{\circ}$ with the horizontal. The trajectory of the body is observed to pass through a point $(20,10)$. If $T$ is the time of flight, then its momentum vector, at time $t =\frac{ T }{\sqrt{2}}$, is.
$\left[\right.$ Take $\left.g=10 m / s ^{2}\right]$
- [JEE MAIN 2022]
- A
$100 \hat{ i }+(100 \sqrt{2}-200) \hat{ j }$
- B
$100 \sqrt{2} \hat{i}+(100-200 \sqrt{2}) \hat{j}$
- C
$100 \hat{ i }+(100-200 \sqrt{2}) \hat{ j }$
- D
$100 \sqrt{2} \hat{i}+(100 \sqrt{2}-200) \hat{j}$
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- [AIIMS 2000]