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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Motion in Two Dimensions

NTA Abhyas 2020 · Physics · Motion in Two Dimensions

A small body is released from point A of smooth parabolic path y =...

A small body is released from point AA of smooth parabolic path y=x2y = {x}^{2}, where yy is vertical axis and xx is horizontal axis at ground, as shown. The body leaves the surface from point BB. If g=10ms−2g = 10m{s}^{- 2} then what is the value of dd (in mm)?

A small body is released from point A of smooth parabolic path y = x ^ 2 , where y is vertical axis and x is horizontal

Correct answer

2

Step-by-step solution

A small body is released from point A of smooth parabolic path y = x ^ 2 , where y is vertical axis and x is horizontal

H1=(2)2=2m{H}_{1} = ( \sqrt{2}{)}^{2} = 2m
H2=(1)2=1m{H}_{2} = ( 1{)}^{2} = 1m
∴v=2×g×(H2−H1)∴ v = \sqrt{2 \times g \times \left({H}_{2} - {H}_{1}\right)}
v=20ms−1v = \sqrt{20}m{s}^{- 1}
tan \theta = {\left\dfrac{dy}{\text{dx}}\right|}_{x = 1} = {\left2x\right|}_{x = 1} = 2
−H2=dtanθ−gd22u2cos⁡2θ- {H}_{2} = dtan \theta - \dfrac{g{d}^{2}}{2{u}^{2}\cos^{2} \theta}

−1=2d−10d22(20)2(15)2- 1 = 2d - \dfrac{10{d}^{2}}{2{\left(\sqrt{20}\right)}^{2}{\left(\dfrac{1}{\sqrt{5}}\right)}^{2}}

54d2−2d−1=0\dfrac{5}{4}{d}^{2} - 2d - 1 = 0

On solving we get d=2md = 2m

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