The world record for the discus throw is 74.08m , set by Jürgen Schult in 1986.If he had been competing, not on earth, but on Mars, where the acceleration due to gravity is 0.379 what it is on earth, and if he had thrown the discus in exactly the same way as on earth, what would be his Martian record for this throw? Assume that the discus is released essentially at ground level

Already answered elsewhere

To find out the Martian record for Jürgen Schult's discus throw, we can use the concept of projectile motion and apply the physics equations related to a projectile's range.

The range of a projectile is the horizontal distance covered during its motion. The formula for the range (R) is given by:

R = (v^2 * sin(2θ)) / g

Where:
- v is the initial velocity of the discus
- θ is the launch angle of the throw
- g is the acceleration due to gravity

Before we proceed, let's convert the given Mars acceleration due to gravity (0.379 times Earth's gravity) into a usable value. Earth's acceleration due to gravity is approximately 9.8 m/s², so for Mars:

g_mars = 0.379 * 9.8 m/s²

Now, let's substitute the values into the range formula:

R_mars = (v^2 * sin(2θ)) / g_mars

We need to assume that the initial velocity and launch angle remain the same. Therefore, the only change required will be in the Martian acceleration due to gravity.

Using Jürgen Schult's world record distance on Earth (74.08m), we can rearrange the equation to solve for v:

v = sqrt((R_earth * g) / sin(2θ))

Now, substitute the Martian gravity value and solve for the Martian range:

R_mars = (v^2 * sin(2θ)) / g_mars

By performing these calculations, we can find the Martian record for Jürgen Schult's discus throw.