Question

The range of a projectile depends not only on v0 and θ0 but also on the value g of 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 g = 9.8128 m/s2). Assuming the same values of v0 and θ0, by how much would his record have differed if he had competed instead in 1956 at Melbourne (where g = 9.7999 m/s2)?

Answer 1

Answer:

0.0106491902979 m

Explanation:

$v_0$ = Velocity of jump

$\theta_0$ = Angle of jump

g = Acceleration due to gravity

Jump at Berlin

$R_1=\dfrac{v_0sin2\theta_0}{g_1}=8.09\ m$

Jump at Melbourne

$R_2=\dfrac{v_0sin2\theta_0}{g_2}$

The difference is

$\Delta R=R_2-R_1\\\Rightarrow \Delta R=\dfrac{v_0sin2\theta_0}{g_2}-\dfrac{v_0sin2\theta_0}{g_1}\\\Rightarrow \Delta R=v_0sin2\theta_0(\dfrac{1}{g_2}-\dfrac{1}{g_1})\\\Rightarrow \Delta R=v_0sin2\theta_0(\dfrac{1}{g_2}-\dfrac{1}{g_1})\\\Rightarrow \Delta R=\dfrac{v_0sin2\theta_0}{g_1}(\dfrac{g1}{g2}-1)\\\Rightarrow \Delta R=R_1(\dfrac{g1}{g2}-1)\\\Rightarrow \Delta R=8.09(\dfrac{9.8128}{9.7999}-1)\\\Rightarrow \Delta R=0.0106491902979\ m$

The record would have differed by 0.0106491902979 m