Question

A human expedition lands on an exoplanet. One of the explorers is able to jump a maximum distance of 19.0 m with an initial speed of 2.40 m/s. Find the gravitational acceleration on the surface of the exoplanet. Assume the planet has a negligible atmosphere. (Enter the magnitude in m/s2.)

Answer 1

0.303 m/s²

Step-by-step explanation:

Maximum range for a project is at 45°.

Given:

sₓ = 19.0 m

uₓ = 2.40 cos 45° = 1.70 m/s

aₓ = 0 m/s²

sᵧ = 0 m

uᵧ = 2.40 sin 45° = 1.70 m/s

Find: aᵧ

First solve for time.

sₓ = uₓ t + ½ aₓ t²

19.0 = 1.70 t + ½ (0) t²

t = 11.2 s

Next, solve for acceleration.

sᵧ = uᵧ t + ½ aᵧ t²

0 = 1.70 (11.2) + ½ aᵧ (11.2)²

aᵧ = -0.303 m/s²

Alternatively, you can use the range equation:

R = u² sin(2θ) / g

19.0 = (2.40)² / g

g = 0.303 m/s²