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Standard XII

Physics

Summary for Time, Height and Range

A body is pro...

Question

A body is projected at $t = 0$ with a velocity $10 m s^{1}$ at an angle of $60$ with the horizontal.The radius of curvature of its trajectory at $t = 1 s$ is $R$ . Neglecting air resistance and taking acceleration due to gravity $g = 10 m s 2$ , the value of $R$ is :

2.4 m

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10.3 m

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2.8 m

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5.1 m

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Solution

The correct option is C $2.8 m$
$v_{x} = 10 cos 60^{o} = 5 m / s$
$v_{y} = 10 cos 30^{o} = 5 \sqrt{3} m / s$
Velocity after $t = 1$ sec
$v_{x} = 5 m / s$
$v_{y} = | (5 \sqrt{3} - 10) | m / s = 10 - 5 \sqrt{3}$
$a_{n} = \frac{v^{2}}{R} \Rightarrow \frac{v_{x}^{2} + v_{y}^{2}}{a_{n}} = \frac{25 + 100 + 75 - 100 \sqrt{3}}{10 cos θ}$
$tan θ = \frac{10 - 5 \sqrt{3}}{5} = 2 - \sqrt{3} \Rightarrow θ = 15^{o}$
$R = \frac{100 (2 - \sqrt{3})}{10 cos 15} = 2.8 m$

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A body is projected at t=0 with a velocity 10ms1 at an angle of 60 with the horizontal.The radius of curvature of its trajectory at t=1s is R. Neglecting air resistance and taking acceleration due to gravity g=10ms2, the value of R is :

The correct option is C 2.8mvx=10cos60o=5m/svy=10cos30o=5√3m/sVelocity after t=1 secvx=5m/svy=|(5√3−10)|m/s=10−5√3an=v2R⇒v2x+v2yan=25+100+75−100√310cosθtanθ=10−5√35=2−√3⇒θ=15oR=100(2−√3)10cos15=2.8m

A body is projected at $t = 0$ with a velocity $10 m s^{1}$ at an angle of $60$ with the horizontal.The radius of curvature of its trajectory at $t = 1 s$ is $R$ . Neglecting air resistance and taking acceleration due to gravity $g = 10 m s 2$ , the value of $R$ is :