Question

A particle projected vertically up from the ground, is
found to be at a height h, when t=4 s and again at
t=6 s. If acceleration due to gravity is , the
value of h is

Answer 1

Answer:

this is required answer

Answer 2

Answer:

  Acceleration due to gravity is , the value of h is 12g

Explanation:

Given :

      A particle projected vertically up from the ground .

      Height at h,

    Found to be at a height h, when t=4 s and again at t=6 s

We know that $S=ut+\frac{1}{2} at^{2}$

               $a=-g,S=h$

               $h=ut-\frac{1}{2} gt^{2}$    

                 $\frac{1}{2} gt^{2}-ut+h$

                This is of the form     $ax^{2} +bx+c=0$

                          The sum of the roots = $\frac{-b}{a}$

                           The product of the roots $\frac{c}{a}$

                        $10=\frac{2u}{g}$

                        $u=5g$

                     $\frac{1}{2} gt^{2}-ut+h=0$

                      $6*4=\frac{2h}{g}$

                   h=12g  

     

                Acceleration due to gravity is , the value of h is 12g

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