Question

A particle projected from ground moves at angle 45º with horizontal one second after projection and speed is minimum two seconds after the projection. the angle of projection of particle is

Answer 1

X=u cosθt   and Vx= u cosθY=u sinθt -12gt2and Vy=usinθ  -gtThe horizontal component of speed is Vx is always constant .So the speed is minimum when the vertical component Vy=0.So , we get , u sinθ=2g ....(1)Direction of velocity of the projecile at time t=1 sec = 45°Then , tan45°=1=VyVx=usinθ-gu cosθ⇒u cosθ=g ......(2)So dividing equation 1 and 2 we get,tan θ=2⇒θ=tan-12
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Answer 2

$X = u cos θ t \\ V_{x} = u cos θ \\ </p><p>y = u sin θ t - \frac{g {t}^{2} }{2} \\ V_{y} = u sin θ - g t</p><p></p><p>$

The horizontal component of speed ${V_x}$ is always constant.

So , the speed is minimum when the vertical component ${V_y}$  is 0.

$\longrightarrow$ u sinθ = 2 g 

45° = direction of velocity of the project at  time t = 1 sec.

 

$tan \: 45° = 1 = \frac{ v_{x} }{ v_{y} } \\ </p><p> u cosθ = g</p><p></p><p>  tan θ = 2     \\ \\  θ = tan⁻¹ 2$