A person can jump a horizontal distance of
2.59 m on the Earth.
The acceleration of gravity is 11.8 m/s2 .
How far could he jump on the Moon, where
the free-fall acceleration is 0.297g?
Answer in units of m
To find out how far the person could jump on the Moon, we can make use of the concept of projectile motion. The horizontal distance covered by a projectile can be calculated by analyzing its vertical and horizontal motion independently.
On Earth:
1. Given vertical motion: The acceleration of gravity on Earth is 9.8 m/s^2. We have no information about the time or vertical distance, so for simplicity, let's assume the person jumps straight up and lands at the same level.
2. Using kinematic equation: The equation we can use is d = v_i*t + (1/2)*a*t^2, where d is the vertical distance, v_i is the initial vertical velocity, a is the acceleration, and t is the time.
For upward motion: The initial vertical velocity is 0 m/s since the person jumps straight up. The acceleration, a, is -9.8 m/s^2 (negative because it is in the opposite direction of the motion) and the time, t, is the time it takes for the person to reach the maximum height and come back down.
Since the person lands at the same level, the vertical displacement, d, is 0.
So we can rewrite the equation as 0 = 0 + (1/2)*(-9.8)*t^2.
Solving for t, we find t = sqrt(0) = 0. This means the person takes no time to reach the maximum height and is instantly back at the same level.
3. Given horizontal motion: The horizontal distance covered is 2.59 m, assuming no air resistance.
Since there is no acceleration in the horizontal direction, we can use the equation d = v*t, where d is the horizontal distance, v is the horizontal velocity, and t is the time.
Since the person jumps straight up, the horizontal velocity is constant throughout the motion.
Thus, d = v*t = v*0 = 0.
Therefore, the person does not cover any horizontal distance while jumping on Earth.
On the Moon:
4. Given vertical motion: The free-fall acceleration on the Moon is 0.297g, where g is the acceleration due to gravity on Earth (9.8 m/s^2).
Using the same equation as in step 2, d = v_i*t + (1/2)*a*t^2, but with new values, we can calculate the time taken to reach the maximum height and come back down.
For upward motion: The acceleration, a, is -0.297g and the vertical displacement, d, is 0.
So we have 0 = 0 + (1/2)*(-0.297*9.8)*t^2.
Solving for t, we find t = sqrt(0) = 0. The time taken for vertical motion on the Moon is also 0.
5. Given horizontal motion: Assuming no air resistance, the horizontal distance covered on the Moon can be calculated using the equation d = v*t, where d is the horizontal distance, v is the horizontal velocity, and t is the time.
Since the person jumps straight up, the horizontal velocity is constant throughout the motion.
Using the vertical motion equation's time (0), we can calculate the horizontal distance.
d = v*t = v*0 = 0.
Thus, the person does not cover any horizontal distance while jumping on the Moon.
Therefore, the person cannot jump any horizontal distance on both Earth and the Moon, as determined by the given information.