91Ó°ÊÓ

Understanding projectile motion requires the use of kinematic equations, which describe the motion of objects under the influence of gravity. These equations allow us to calculate parameters like time of flight, initial, and final velocities, and displacement. In projectile motion, motion is typically separated into horizontal and vertical components. The time of flight and velocity aspects are handled through these kinematic equations:

• Vertical motion: - The equation: \( y = v_{iy} t + \frac{1}{2} a t^2 \) describes vertical displacement, where \(y\) is the height, \(v_{iy}\) is the initial vertical velocity, \(a\) is acceleration due to gravity, and \(t\) is time.
• Horizontal motion: - Velocity is constant, so: \( x = v_{ix} t \), where \(x\) is horizontal distance and \(v_{ix}\) is horizontal velocity.

By using these equations, we determine how long a projectile stays in air, how far it travels, and how its velocity changes over time.

Initial velocity is a critical component in projectile motion, as it determines the projectile's trajectory. It is composed of two parts: the horizontal and vertical components.

• Horizontal component: - Calculated as \( v_{ix} = v_i \cos(\theta) \), where \( v_i \) is the speed and \( \theta \) is launch angle.
• Vertical component: - Calculated as \( v_{iy} = v_i \sin(\theta) \).

For the shot-put exercise, the initial velocity at a speed of 12 m/s and angle of 20° results in a vertical velocity of approximately 4.10 m/s and horizontal velocity of 11.28 m/s. This breakdown helps analyze the separate influences of these components on the projectile's overall motion.

Time of flight refers to the total time a projectile remains in the air. To calculate it accurately for an elevated launch, you'll need to consider both the initial height and initial vertical velocity. For a projectile launched from a higher point, like the 2 meters elevation in our exercise, the time of flight will differ from launching at ground level. We'll apply vertical kinematic equations, solving:\[ 0 = y = v_{iy} t + \frac{1}{2} (-g) t^2 + y_0 \]\[ 0 = 4.10t - 4.9t^2 + 2 \]This is a quadratic equation, and solving it gives the projectile a time of flight of approximately 1.07 seconds compared to 0.84 seconds if launched from the ground level only. This means the elevated launch increases the air time.

The horizontal range is the distance a projectile covers along the horizontal axis. It is influenced by the initial horizontal velocity and the time the projectile is in the air.

• Formula: \[ R = v_{ix} \times t \]
• Here, \(v_{ix}\) is the horizontal component of initial velocity, and \( t \) is the time of flight.

In the exercise, with \( v_{ix} \) approximately 11.28 m/s and a time of flight of 1.07 seconds, the horizontal range is about 12.07 meters. By understanding this calculation, one can predict where a projectile will land, which is essential for optimizing the shot-putter's technique.