Nasa moon vechile is being used to determine an accurate value of acceleration due to gravity on the moon. the vechile is remote contrlled and positioed so that it is stationary and on level ground. the projectile is fired vertically upwards from the deck of the vechile and it reaches a height of 3.6m before landing back on vechile. the time of flight is 4.26s. there is no atmosphere on the moon.

1. show intinal velocity of projectile is 3.4s
2. Hence or otherwise determine the accerlation due to gravity on the moon.
3. The time taken for the projectile to reach a height of 2.4m in 0.90s. write down one other time when the projectile is at this height.

To determine the acceleration due to gravity on the Moon using the information provided, we can apply some fundamental physics principles. The projectile's motion can be analyzed using the equations of motion, particularly since we know the initial velocity, maximum height, and total time of flight. Let's break this down step by step.

Understanding the Motion of the Projectile

When a projectile is fired vertically upwards, it experiences a constant acceleration due to gravity, which acts downwards. On the Moon, this acceleration is what we are trying to find. The key points to note are:

• The initial velocity of the projectile is given as 3.4 m/s.
• The maximum height reached is 3.6 m.
• The total time of flight is 4.26 s.

Using the Kinematic Equation

We can use the following kinematic equation to relate the initial velocity (u), acceleration (a), time (t), and displacement (s):

s = ut + (1/2)at²

For the upward motion to the maximum height, the displacement (s) is 3.6 m, the initial velocity (u) is 3.4 m/s, and we need to find the acceleration (a), which will be negative since it acts downwards. The time taken to reach the maximum height can be calculated as half of the total time of flight, which is:

t = 4.26 s / 2 = 2.13 s

Calculating the Acceleration

Now, substituting the values into the kinematic equation:

3.6 m = (3.4 m/s)(2.13 s) + (1/2)(a)(2.13 s)²

Calculating the first term:

3.6 m = 7.242 m + (1/2)(a)(4.5369 s²)

Rearranging gives:

3.6 m - 7.242 m = (1/2)(a)(4.5369 s²)

-3.642 m = (1/2)(a)(4.5369 s²)

Now, multiplying both sides by 2:

-7.284 m = a(4.5369 s²)

Solving for a:

a = -7.284 m / 4.5369 s² ≈ -1.61 m/s²

This value represents the acceleration due to gravity on the Moon, which is approximately -1.61 m/s². The negative sign indicates that the acceleration is directed downwards.

Finding the Time at a Specific Height

Next, we need to determine when the projectile is at a height of 2.4 m. We can use the same kinematic equation:

2.4 m = (3.4 m/s)(t) + (1/2)(-1.61 m/s²)(t²)

Rearranging gives us a quadratic equation:

0 = -0.8055t² + 3.4t - 2.4

Using the quadratic formula, t = [ -b ± √(b² - 4ac) ] / 2a, where:

• a = -0.8055
• b = 3.4
• c = -2.4

Calculating the discriminant:

D = (3.4)² - 4(-0.8055)(-2.4)

D = 11.56 - 7.728 = 3.832

Now substituting into the quadratic formula:

t = [ -3.4 ± √(3.832) ] / (2 * -0.8055)

Calculating the two possible times:

t₁ = [ -3.4 + 1.96 ] / -1.611 = 0.90 s

t₂ = [ -3.4 - 1.96 ] / -1.611 = 2.73 s

Thus, the projectile is at a height of 2.4 m at two different times: 0.90 seconds on the way up and approximately 2.73 seconds on the way down.

Summary

In summary, we determined that the acceleration due to gravity on the Moon is approximately -1.61 m/s². Additionally, the projectile reaches a height of 2.4 m at 0.90 seconds and again at about 2.73 seconds during its flight. This analysis illustrates the principles of projectile motion and the effects of gravity in a vacuum, such as on the Moon.