91影视

It is given that,

$$\begin{gathered} R_1=8.09\mathrm{m} \\ {g}_1=9.8128\mathrm{m}/{\mathrm{s}}^2 \\ {g}_2=9.7999\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The problem deals with the range of the projectile which is the displacement in the horizontal direction. Thus, using the equation for the range of the projectile, the jump length can be computed using two different values of g and further the difference can be found.

Formula:

The range of the projectile is given by,

$R=\frac{v_0^2\sin 2{胃}_0}{g}$ (i)

Where, $v_0$is initial velocity, ${胃}_0$is an angle made with horizontal, g is an acceleration due to gravity.

The jump of the player is an example of projectile motion.

The range of a projectile is given by equation (i),

This indicates that,$R鈭�\frac{1}{g}$

Now,

$$\begin{gathered} R_1=8.09\mathrm{m} \\ {g}_1=9.8128\mathrm{m}/{\mathrm{s}}^2 \\ {\mathrm{g}}_2=9.7999\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Now, calculate R₂ first,

$$\begin{gathered} R鈭�\frac{1}{g}鈬�\frac{R_2}{R_1}=\frac{g_1}{g_2} \\ 鈬�R_2=\frac{R_1脳g_1}{g_2} \\ =\frac{8.09脳9.1828}{9.7999} \\ =8.10\mathrm{m} \end{gathered}$$

Thus, change in the value of R is, 8.10- 8.09 = 0.01 m