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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 ball is thrown with a speed u , at an angle θ with the...

A ball is thrown with a speed uu, at an angle θ\theta with the horizontal. At the highest point of its motion, the strength of gravity is somehow doubled. Taking this change into account, the total time of flight of the projectile isA ball is thrown with a speed u , at an angle θ with the horizontal. At the highest point of its motion, the strength

Options

  1. A2usinθg\dfrac{2usin\theta}{g}
  2. B32usinθg\dfrac{3}{2}\dfrac{usin\theta}{g}
  3. C34usinθg\dfrac{3}{4}\dfrac{usin\theta}{g}
  4. D(2+12)usinθg\left(\dfrac{\sqrt{2} + 1}{\sqrt{2}}\right)\dfrac{usin\theta}{g}

Correct answer

D. (2+12)usinθg\left(\dfrac{\sqrt{2} + 1}{\sqrt{2}}\right)\dfrac{usin\theta}{g}

Step-by-step solution

A ball is thrown with a speed u , at an angle θ with the horizontal. At the highest point of its motion, the strength

Time of ascent

t1=usinθg{t}_{1} = \dfrac{usin\theta}{g}

Time of descent

t2=2H2g{t}_{2} = \sqrt{\dfrac{2H}{2g}}

t2=u2sin2θ2g2=usinθ2g{t}_{2} = \sqrt{\dfrac{u^{2}sin^{2}\theta}{2g^{2}}} = \dfrac{usin\theta}{\sqrt{2}g}

t=t1+t2=usinθg+usinθ2gt = {t}_{1} + {t}_{2} = \dfrac{usin\theta}{g} + \dfrac{usin\theta}{\sqrt{2}g}

⇒ \dfrac{usin\theta}{g} \left(\dfrac{\sqrt{2} + 1}{\sqrt{2}}\right)
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