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Motion Of Body Under Gravity - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • 20 Questions around this concept.

Solve by difficulty

A ball is released from a height \mathrm{h}. If  \mathrm{t_{1} \: and \: t_{2}}  be the time required to complete the first half and second half of the distance respectively. Then, choose the correct relation between  \mathrm{ t_{1} \: and \: t_{2}}.

 

Concepts Covered - 1

Motion of Body Under Gravity (Free Fall)

The force of attraction of the earth on anybody is called the force of gravity. Acceleration produced on the body by the force of gravity is called acceleration due to gravity. It is represented by the symbol ‘g’.

Sign convention:

  • Upward direction and right direction is taken as positive
  • Downward direction and left direction is taken as negative

There are three cases basically in this - 

             1) If a body dropped from some height (initial velocity zero)

                      u = 0 

                      a = g

                     \begin{array}{l}{v=g t} \\ {h=\frac{1}{2} g t^{2}} \\ {v^{2}=2 g h} \\ {h_{n}=\frac{g}{2}(2 n-1)}\end{array}

            2) If a body is projected vertically downward with some initial velocity

                   \begin{array}{l}{\text { Equation of motion: } \quad v=u+g t} \\ {\qquad \begin{array}{ll}{h=} & {u t+\frac{1}{2} g t^{2}} \\ {v^{2}} & {=u^{2}+2 g h} \\ {h_{n}} & {=u+\frac{g}{2}(2 n-1)}\end{array}}\end{array}

      

            3) If a body is projected vertically upward.

                 \dpi{120} \\\mathrm{\left(i\right)\:Apply\:equation\:of\:motion:}\\\mathrm{Take\:initial\:position\:as\:origin\:and\:the\:direction\:of\:motion\:\left(vertically\:up\right)\:as}\\\mathrm{a=-g\;\;\left[as\:acceleration \ due \ to \ gravity\:is\:downwards\right]}\\\mathrm{So,\:if\:the\:body\:is\:projected\:with\:velocity}\;\;u,\\\mathrm{\text{\:}\:and\:after\:time\:\:t\:\text{\:}\:it\:reaches\:up\:to\:height\:\:h\:\text{\:}\:then\:},\\\mathrm{v=u-g\:t\:;\:\quad \:h=u\:t-\frac{1}{2}\:g\:t^2\:;\:v^2=u^2-2\:g\:h}

              \begin{array}{l}{\text { (ii) For the case of maximum height } v=0} \\ {\text { So from the above equation }} \\ {\qquad \begin{array}{c}{u=g t} \\ {h=\frac{1}{2} g t^{2}} \\ {\text { and } u^{2}=2 g h}\end{array}}\end{array}  

 

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Motion of Body Under Gravity (Free Fall)

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