91Ó°ÊÓ

Kinematic equations are essential in analyzing the motion of objects, especially in projectile motion where the object moves both vertically and horizontally. These equations relate displacement, initial velocity, final velocity, acceleration, and time. In the case of a ball thrown vertically, like in our problem, the important equation is:\[ s(t) = v_0 t - \frac{1}{2} g t^2 \]Here, \( s(t) \) represents the height of the ball at time \( t \), \( v_0 \) is the initial velocity, and \( g \) is the acceleration due to gravity, which is \( 9.8 \text{ m/s}^2 \) downward. This equation helps us determine how high the ball goes after a certain time, taking into account that gravity is constantly decelerating it on the way up and accelerating it on the way down. The understanding of this equation lays the foundation for further calculations like initial velocity and height in vertical motion problems.

Calculating the initial velocity is a critical step in solving projectile motion problems. It is the speed at which the object is launched into the air. In our example, the ball passes the overhead power line at two different times, once on its way up and then again on its way down. By setting up equations for the height reached at these two times, we can solve for the initial velocity.From the equation \( h = v_0 t - \frac{1}{2} g t^2 \), we have two instances:

• At \( t_1 = 0.75 \) seconds, the equation is \( h = v_0 (0.75) - \frac{1}{2} (9.8)(0.75)^2 \).
• At \( t_2 = 1.5 \) seconds, it becomes \( h = v_0 (1.5) - \frac{1}{2} (9.8)(1.5)^2 \).

By subtracting these equations, we isolate \( v_0 \), which allows us to solve for this unknown:\[ 0 = v_0 (0.75) - 9.8 \times 1.125 \]Solving this gives \( v_0 = 14.7 \, \text{m/s} \). This calculation helps us understand the speed needed for the ball to reach the power line at the given times.

Vertical motion analysis involves understanding how an object moves under the influence of gravity, typically upwards and then downwards. Our ball, thrown vertically, passes the same height (the power line) twice: once going up and once coming back down.The ball's height at any time is influenced by its initial velocity and the constant acceleration due to gravity. Given:

• Time up to the power line: \( 0.75 \) s
• Total time back to power line: \( 1.5 \) s

We calculate the height using the kinematic equation:\[ h = v_0 t - \frac{1}{2} g t^2 \]With \( v_0 = 14.7 \, \text{m/s} \), substituting it back into our equation for \( t = 0.75 \):\[ h = 14.7 \times 0.75 - \frac{1}{2} \times 9.8 \times 0.75^2 \]\[ h = 11.025 - 2.75625 = 8.26875 \, \text{m} \]This method shows how we can precisely determine the height of the power line by analyzing the ball's motion and making use of consistent physical laws such as gravity. Understanding vertical motion helps in grasping how objects behave in freefall and projectile settings.