Question

An astronaut on the planet Zircon tosses a rock horizontally with a speed of 6.60 m/s . The rock falls through a vertical distance of 1.50 m and lands a horizontal distance of 8.65 m from the astronaut. What is the size of the acceleration due to gravity on Zircon?
gzircon = m/s2

Answer 1

Answer:

gzircon = 1.75 m/s²

Explanation:

Once released, as no forces are acting on the rock in the horizontal direction, it continues moving at the same speed, until it lands at a horizontal distance of 8.65 m from the astronaut.

Applying the definition of average velocity, and solving for t, we have:

$t = \frac{x}{v} =\frac{8.65m}{6.60 m/s} = 1.31 s (1)$

As the time must be the same for the movement in the vertical direction, when the rock finishes its fall (due only to the influence of the gravity of Zircon), we can write the following equation:

$y = \frac{1}{2} * gz*(t)^{2} = 1.5 m$

Replacing by the value of the time, which is the same that we found in (1), we have:

$y = \frac{1}{2} * gz*(1.31s)^{2} = 1.5 m$

Solving for gzircon, we get:

$gzircon = \frac{2*1.5 m}{(1.31s)^{2} } = 1.75 m/s2$

⇒ gzircon = 1.75 m/s²