Question

1. in a classic experiment, a ball (ball 1) is dropped from a specific height. at the same time, an identical ball (ball 2) is launched horizontally from the same height. which ball hits the ground first? the dropped ball (ball 1) because its path is shorter. the launched ball (ball 2) because the balls velocity is greater. the two balls will reach the ground at the same time because the horizontal and vertical planes of motion are independent of each other and they were also released from the same height. it is not possible to determine which ball hits the ground first because of the complex nature of motion.

Explanation:

Step1: Analyze vertical - motion

Both balls have an initial vertical velocity \(v_{0y}=0\) and are in free - fall with acceleration \(a = g\) (acceleration due to gravity). The vertical displacement \(y - y_0=-h\) (taking downwards as negative). Using the kinematic equation \(y - y_0=v_{0y}t+\frac{1}{2}at^{2}\), we get \(-h = 0\times t-\frac{1}{2}gt^{2}\), so \(t=\sqrt{\frac{2h}{g}}\).

Step2: Analyze horizontal - motion

Ball 2 has a non - zero initial horizontal velocity \(v_{0x}\), but this horizontal motion does not affect the vertical motion. The time of flight is determined only by the vertical motion parameters (height \(h\) and acceleration \(g\)). Since both balls have the same initial vertical conditions and are dropped from the same height \(h\), they will take the same time to reach the ground.

Answer:

The two balls will reach the ground at the same time because the horizontal and vertical planes of motion are independent of each other and they were also released from the same height.