91Ó°ÊÓ

To understand projectile motion, we must rely on kinematic equations. These equations help us analyze the motion of objects in physics when they are in free fall or thrown into the air like a ball. In this problem, the ball's trajectory involves both horizontal and vertical motion. - **Horizontal motion**: Here, velocity remains constant because there's no acceleration.- **Vertical motion**: This is affected by gravity, where acceleration due to gravity is denoted by \( g = 9.8 \, \mathrm{m/s^2} \).The equations we can use include:1. The basic kinematic equation for vertical motion: \[ h = v_{y0} t + \frac{1}{2} g t^2 \] where \( h \) is the height, \( v_{y0} \) is the initial vertical velocity, and \( g \) is the acceleration due to gravity.2. The horizontal range equation: \[ R = v_{x0} t \] where \( R \) is the range, and \( v_{x0} \) is the initial horizontal velocity. These equations allow us to calculate the horizontal range and the time of flight, given the initial speed and angle of projection.

The horizontal range of a projectile is the total distance it travels horizontally. For the ball to meet the runner at the right spot, both the horizontal range of the ball and the distance the woman runs need to be the same.To find out the range, we need to determine the initial horizontal velocity \( v_{x0} \) using:- The formula: \[ v_{x0} = v_0 \cos \theta \]Next, the horizontal range \( R \) is calculated by multiplying the initial horizontal velocity with the time of flight \( t \):- \[ R = v_{x0} \times t \]Since the woman runs with a speed of 6.00 m/s to catch the ball, her distance covered is equal to the ball’s range when their speeds adjust for time \( t \).This relationship is crucial because it helps us know the exact location where the ball and woman meet, given she maintains a constant speed.

Time of flight is the duration for which the projectile remains in the air. It’s influenced by the initial speed, angle of projection, and the height from which it is thrown. To calculate the time of flight, we solve the vertical motion equation:- \[ h = v_{y0} t + \frac{1}{2} g t^2 \]In our scenario, the height \( h \) is \(- 45\) meters, indicating a downward motion. The initial vertical speed \( v_{y0} \) relates to the angle \( \theta \) through \( v_{y0} = v_0 \sin \theta \).The equation is quadratic in form, and solving it gives us the time \( t \) when the ball reaches the ground level from the cliff’s height.Knowing the time of flight is necessary as it directly affects how far the projectile will travel horizontally, which is equal to the distance the woman runs to catch the ball.