A man on the moon throws a ball vertically upwards and it is notices that the ball travels 3.0m less in the fifth second of its upward motion that it does in the third second. What is the acceleration due to gravity on the moon?
To find the acceleration due to gravity on the moon, we need to analyze the motion of the ball.
Let's denote the initial velocity of the ball as 'u' and the acceleration due to gravity on the moon as 'g'.
In the first second, the ball travels some distance 's1'. In the second second, it travels another distance 's2'. And so on.
From the information given, we know that the ball travels 3.0m less in the fifth second (s5) than it does in the third second (s3).
Using the equations of motion, we can analyze the motion of the ball.
In the fifth second (t5), the ball reaches its maximum height and starts to descend. At this point, its velocity is zero (since it momentarily comes to rest before falling back down).
In the third second (t3), the ball is still moving upwards. Therefore, its velocity at this point is positive. Let's denote the velocity at t3 as 'v3'.
Now, we can set up two equations:
Equation 1: s3 = u * t3 + (0.5) * (-g) * (t3)^2 -> Equation for distance traveled in the third second
Equation 2: s5 = u * t5 + (0.5) * (-g) * (t5)^2 -> Equation for distance traveled in the fifth second
Since the ball is at rest at the maximum height during the fifth second, its velocity in the fifth second is also zero. Therefore, we can assume t5 = 2 * t3 (since the motion is symmetric).
Substituting the value of t5 in Equation 2, we get:
s5 = u * (2 * t3) + (0.5) * (-g) * (2 * t3)^2
Now, we know that s5 is 3.0m less than s3:
s5 = s3 - 3.0
Substituting the values of s3 and s5 in the equations and rearranging, we get:
u * t3 + (0.5) * (-g) * (t3)^2 = u * (2 * t3) + (0.5) * (-g) * (2 * t3)^2 - 3.0
Simplifying the equation further, we get:
-g * (t3)^2 + 2 * g * (t3)^2 - u * t3 + 2 * u * t3 - 3.0 = 0
-g * (t3)^2 + 2 * u * t3 - 3.0 = 0
This is a quadratic equation in terms of 't3'. Solving this equation will give us the value of 't3'. Once we know 't3', we can substitute it back into the equation to find the value of 'g' (acceleration due to gravity on the moon).
Please note that this problem requires solving a quadratic equation, and it involves some calculations. You can either use algebraic methods or use numerical methods to solve the equation.