Question

1. the objects thrown from the ground with initial speeds of v1, v2, and v3, as shown in the figure, travel the horizontal distances of x1, x2, and x3, respectively. what is the relationship between x1, x2, and x3? a) x1 > x2 > x3 b) x3 > x2 > x1 c) x3 > x1 > x2 d) x1 > x2 = x3 e) x1 = x3 > x2

Explanation:

Step1: Analyze the time of flight

The time of flight \( t\) for a projectile (assuming vertical motion) is determined by the vertical component of the motion. Since the objects are thrown from the same height and the vertical acceleration \( g\) is the same for all ( \( y = v_{0y}t+\frac{1}{2}gt^{2}\), and if we assume the initial vertical velocity component \( v_{0y}=0\) for horizontal projection, then \( y=\frac{1}{2}gt^{2}\), \(t = \sqrt{\frac{2y}{g}}\)). So the time of flight \(t\) is the same for all three objects.

Step2: Use the horizontal - distance formula

The horizontal distance \(x\) is given by \(x = v_{x}t\) (where \(v_{x}\) is the initial horizontal velocity and \(t\) is the time of flight). Since \(t\) is constant, and from the figure \(v_{1}

Answer:

B. \(x_{3}>x_{2}>x_{1}\)