Question

The range of a projectile depends not only on ${\mathit{v}}_{\mathbf{0}}$and${\mathit{\theta }}_{\mathbf{0}}$but also on the value gof the free-fall acceleration, which varies from place to place. In 1936, Jesse Owens established a world’s running broad jump record of 8.09 m at the Olympic Games at Berlin (where $\mathit{g}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{8128}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$ ). Assuming the same values of ${\mathit{v}}_{\mathbf{0}}$ and ${\mathit{\theta }}_{\mathbf{0}}$ ,by how much would his record have differed if he had competed instead in 1956 at Melbourne (where $\mathit{g}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{7999}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$)?

Short answer

The change in the value of the jump distance for different values of g is 0.01 m

Step-by-step solution

Given information

It is given that,

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

Determining the concept

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,

$R=\frac{v_0^2\sin 2{\theta }_0}{g}$ (i)

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

(a) Determining the change in the value of the jump distance for different value of g

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

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

This indicates that,$R\propto \frac{1}{g}$

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,

$$\begin{gathered} R\propto \frac{1}{g}\Rightarrow \frac{R_2}{R_1}=\frac{g_1}{g_2} \\ \Rightarrow R_2=\frac{R_1\times g_1}{g_2} \\ =\frac{8.09\times 9.1828}{9.7999} \\ =8.10\mathrm{m} \end{gathered}$$

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