Respuesta :

Answer:

remains constant.

Explanation:

Diagram of freefall :-

[tex]\setlength{\unitlength}{2mm}\begin{picture}(0,0)\thicklines\put(3,6){\circle{3}}\put(2.9,4.5){\vector(0,-3){2cm}}\put(-3.9,-10){\line(3,0){3cm}}\multiput(-3.5,-10)(0.8,0){18}{\line(1,-2){2mm}}\put(-3.7,0){$\sf\footnotesize 9.8 ms^{-2}$}\put(6,0){\vector(0,3){1.2cm}}\put(5.5,6){\line(3,0){2mm}}\put(5,-1.2){\sf\footnotesize height}\put(6,-2){\vector(0,-3){1.2cm}}\put(5.5,-8){\line(1,0){2mm}}\end{picture}[/tex]

If difference of time is 1s .

  • Acceleration due to gravity applied to both is constant and approximately equal to 9.8m/s^2.

As acceleration remains constant .

Velocity of objects would increase in constant manner.

So distance remains same/constant until the first ball hits the ground.

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