Questions: If you happened to be laying on a train that had a velocity of 10 m / s when you tossed the ball straight up, how far would the train have moved while the ball was in the air?

Solution Steps

To find out how long the ball is in the air, we need to consider the motion of the ball under gravity. When you toss the ball straight up, it will rise to a certain height and then fall back down. The total time the ball is in the air is twice the time it takes to reach the highest point.

Using the kinematic equation for the upward motion: \[ v_f = v_i + at \] where:

• \( v_f \) is the final velocity (0 m/s at the highest point),
• \( v_i \) is the initial velocity (let's denote it as \( v_0 \)),
• \( a \) is the acceleration due to gravity (-9.8 m/s\(^2\)),
• \( t \) is the time to reach the highest point.

Setting \( v_f = 0 \): \[ 0 = v_0 - 9.8t \] \[ t = \frac{v_0}{9.8} \]

The total time the ball is in the air is: \[ T = 2t = 2 \left( \frac{v_0}{9.8} \right) = \frac{2v_0}{9.8} \]

The train is moving at a constant velocity of 10 m/s. The distance the train moves while the ball is in the air can be calculated using the formula: \[ \text{Distance} = \text{Velocity} \times \text{Time} \]

Substituting the total time \( T \): \[ \text{Distance} = 10 \times \frac{2v_0}{9.8} = \frac{20v_0}{9.8} \]

To simplify the expression, we can divide 20 by 9.8: \[ \text{Distance} = \frac{20v_0}{9.8} \approx 2.0408v_0 \]

Final Answer