Question

Suppose an object is thrown with the same initial velocity and an angle theta on the earth and on the moon if the gravitational acceleration of the moon is 1/6 as large as it on the earth what is the maximum height of the projection on the moon?

Answer 1

To find the maximum height of the projection on the moon, use the equation for projectile motion and account for the reduced acceleration due to gravity on the moon. Split the initial velocity into horizontal and vertical components, determine the time taken to reach the maximum height, and use the equation for displacement to find the maximum height on the moon.

Step-by-step explanation:

To find the maximum height of the projection on the moon, we can use the equation for projectile motion. The maximum height is reached when the vertical component of velocity becomes zero (at the highest point of the trajectory). Given that the gravitational acceleration on the moon is 1/6 of that on Earth, we can use the same initial velocity and angle as on Earth, but with the reduced acceleration of gravity on the moon.

Let's assume the initial velocity as 'v' and the initial angle as 'θ'. We can split this initial velocity into horizontal and vertical components: vx = v * cos(θ) and vy = v * sin(θ).

Using the equation vy = uy + at (where uy is the initial vertical velocity and 'a' is the acceleration due to gravity), we can determine the time it takes for the object to reach the maximum height.

At the highest point, the vertical component of velocity becomes zero, so we can use the equation vy = uy - gt (where g is the acceleration due to gravity on the moon). Solving these equations, we can find the time taken to reach the maximum height.

Finally, we can use the equation s = uyt - (1/2)gt2 (where s is the maximum height) to calculate the maximum height on the moon by substituting the known values.