Question

The Apollo 15 mission landed on the moon in 1971. The astronaut David Scott dropped a hammer and a feather. They were released from rest at the same time from the same height. The hammer and the feather landed at the same time. The graph shows how the velocity of the hammer changed with time. a. Use the graph to calculate acceleration due to gravity on the moon. Give the unit. b. Use the graph to calculate the height the hammer was dropped from

Answer 1

Answer:

a. The acceleration due to gravity on the Moon is approximately 1.6 ft/s²

b. The height from which the hammer was dropped is 1.253 feet

Explanation:

a. Whereby the unit of velocity is in ft/s, we have

Acceleration = Rate of change of velocity with time

Therefore, acceleration, a = Slope of the velocity to time graph

Whereby the graph is a straight line graph, we proceed to find the slope as follows;

We consider two points, (x₁, y₁) and (x₂, y₂), where (x₁, y₁) is the origin as follows

x₁ = 0 seconds

x₂ = 0.94 seconds

y₁ = 1.5 ft/s

y₂ = 0 ft/s

The two points are therefore;

(0, 0) and (0.94, 1.5)

The slope of the graph, a is given as follows;

$Slope\, of \, the\, graph = a = \dfrac{y_{2}-y_{1}}{t_{2}-t_{1}} = \dfrac{1.5-0}{0.94-0} = 1.596 \, ft/s^{2}$

The acceleration due to gravity on the Moon = 1.596 ft/s² ≈ 1.6 ft/s²

The unit of the acceleration is ft/s²

b. The height, h, from which the hammer was dropped is found using the following relation;

v² = u² + 2×a×h

Where:

v = Final velocity = Maximum value of the velocity = 2.0 ft/s

u = Initial velocity = 0 ft/s for an object starting from rest

a = Acceleration due to gravity ≈ 1.6 ft/s²

Therefore;

h = (v² - u²)/(2×a) = (2.0² - 0²)/(2×1.6) = 1.253 ft

The height, h, from which the hammer was dropped = 1.253 ft