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Physics Question on projectile motion

A body of mass 10 kg is projected at an angle of 45° 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 = T/√2, is _________.
[Take g = 10m/s2]

A

100i^+(1002−200)j^100\hat{i} + (100\sqrt2-200)\hat{j}

B

1002i^+(100−2002)j^100\sqrt2\hat{i}+(100-200\sqrt2)\hat{j}

C

100i^+(100−2002)j^100\hat{i}+(100-200\sqrt2)\hat{j}

D

1002i^+(1002−200)j^100\sqrt2\hat{i}+(100\sqrt2-200)\hat{j}

Answer

1002i^+(1002−200)j^100\sqrt2\hat{i}+(100\sqrt2-200)\hat{j}

Explanation

Solution

The correct answer is (D):
m = 10 kg
θ=45ºθ = 45º
y=xtan⁡θ(1−xR)y = x \tanθ(1-\frac{x}{R})
⇒10=20(1−20R)⇒ 10 = 20(1-\frac{20}{R})
⇒R=40⇒ R = 40
40=u210⇒u=2040 = \frac{u^2}{10} ⇒ u = 20
⇒T=2×20×12=42S⇒t=2S⇒ T = \frac{2×20×1}{\sqrt2} = \frac{4}{\sqrt2} S ⇒ t = 2S
at t=2,v⃗=(102i^)+(102−2×10)j^t = 2, \vec{v} = (10\sqrt2\hat{i})+(10\sqrt2-2×10)\hat{j}
⇒p→=10[102i^+(102−20)j^]⇒ p→ = 10 [10\sqrt2\hat{i}+(10\sqrt2-20)\hat{j}]
=1002i^+(1002−200)j^= 100\sqrt2\hat{i}+(100\sqrt2-200)\hat{j}