Final answer:
To find the kinetic energy ratio between the highest point and the initial position of a projectile, we analyze the kinetic energy due to horizontal velocity at the highest point and initial total velocity. The ratio is determined by cos²(theta) of the launch angle, leading to a value of 1/4 for a 60° launch angle.
Step-by-step explanation:
The question you've asked relates to kinetic energy and projectile motion in Physics. To find the ratio between the kinetic energy at the highest point and the initial position for a shell fired at an angle, we need to consider the kinetic energy formula and the fact that the vertical component of the velocity becomes zero at the highest point of the projectile's trajectory.
The kinetic energy (KE) of an object is given by KE = (1/2)mv², where 'm' is the mass of the object, and 'v' is its velocity. When the projectile is at its highest point, its vertical component of velocity is zero, and only the horizontal component contributes to the kinetic energy. Initially, the velocity has both horizontal and vertical components. Using the initial speed (u) and the angle to the horizontal (theta), we get the initial vertical and horizontal velocities as u*sin(theta) and u*cos(theta), respectively.
At the highest point, the velocity is only u*cos(theta). So, the kinetic energy at the highest point KEtop = (1/2)m(u*cos(theta))². The initial kinetic energy KEinitial = (1/2)m(u)². Therefore, the ratio of kinetic energies is KEtop / KEinitial = (u*cos(theta))² / (u)² = cos²(theta). In your case, plugging in the given values, with theta being 60°, gives us the ratio cos²(60°) = (1/2)² = 1/4.