91Ó°ÊÓ

Initial kinetic energy is the energy that an object possesses due to its motion at the start of an event. In this exercise, the projectile has an initial kinetic energy of 1550 Joules, which is used to determine the total launch velocity of the object. To find the total initial velocity, we employ the kinetic energy formula:

• \( KE = \frac{1}{2} m v^2 \)

where \( KE \) is kinetic energy, \( m \) is mass, and \( v \) is velocity. By substituting the known values, \( KE = 1550 \) J and \( m = 0.55 \) kg, we solve for velocity. Rearranging the formula gives us \[ v = \sqrt{\frac{2 \, \times \, 1550}{0.55}} \approx 75.3 \text{ m/s} \].
This velocity represents the total speed at which the projectile was launched off the cliff.

When an object is launched into the air, its velocity can be broken down into two components: horizontal and vertical. For this projectile, we need to determine these components to understand how it moves through space. We begin with identifying the vertical component using the formula:

• \( v_y = \sqrt{2gy} \)

where \( y \) is the maximum vertical displacement (140 m) and \( g \) is the acceleration due to gravity (9.8 m/s²). Solving, we find \( v_y \approx 52.4 \text{ m/s} \).
Next, we apply the Pythagorean theorem to find the horizontal component, since velocity in physics can be expressed as a right triangle:

• \( v^2 = v_x^2 + v_y^2 \)

Given \( v = 75.3 \) m/s and \( v_y = 52.4 \) m/s, substituting and solving yields \( v_x \approx 52.7 \text{ m/s} \).
These components illustrate the initial trajectory of the projectile—upwards and outwards.

Vertical displacement refers to the change in height of the projectile relative to its starting point. At maximum height, we calculated this displacement to be 140 m. Importantly, the displacement changes dynamically as the projectile travels. The exercise also explores displacement when the vertical component of velocity is 65 m/s downward. To find the time at which this occurs, we use:

• \( v_y = v_{iy} - g t \)

Substitute \( v_y = 65 \text{ m/s}, v_{iy} = 52.4 \text{ m/s}, \) and solve for \( t \), obtaining approximately 1.28 seconds. To find the displacement at this time, apply:

• \( y = v_{iy} t - \frac{1}{2} g t^2 \)

This gives us \( y \approx 17.1 \text{ m}\), indicating the projectile has risen 17.1 m beyond its starting height at this point.

The Pythagorean theorem is often used in physics to relate the components of vector quantities, such as velocity. It helps in breaking down and understanding the relationship between these components. In this projectile problem, the theorem assists in solving for the horizontal component of the initial velocity using the equation:

• \( v^2 = v_x^2 + v_y^2 \)

The horizontal component, \( v_x \), and vertical component, \( v_y \), form the legs of a right triangle, with the total velocity \( v \) as the hypotenuse. Given the total velocity from the kinetic energy calculation and the vertical velocity at maximum height, we can safely solve for the horizontal velocity. This application underscores the theorem’s utility in simplifying calculations involving vectors, ensuring precision in physics problems.