A projectile is launched at 45° to the horizontal with a force of initial kinetic energy "E". Assuming air resistance to be negligible, what will be the kinetic energy of the project when it reaches its highest point?

Explanation

The following is the solution: When a projectile is launched at an angle of 45 degrees to the horizontal, its horizontal and vertical components of velocity are equal. So, the horizontal velocity (Vx) of the projectile at the highest point is the same as the initial velocity (Vi) multiplied by the cosine of the launch angle: Vx = Vi cos(45 degrees) Since we are considering only the horizontal component of velocity at the highest point, the kinetic energy (K.E-highest) at the highest point can be calculated as follows: K.E-highest = (1/2) x m x Vx2 Substituting the expression for Vx: K.E-highest = (1/2) x m x (Vi cos(45degrees))2 Now, cos(45 degrees) is equal to √2/2: K.E-highest = (1/2) x m x (Vi √2 /2)2 = (1/2) x m x (Vi2 (2 /4)) = (1/2) x m x (Vi2 / 2) Finally, since the initial kinetic energy (Ki) is given as "i," we can write the equation as: K.E-highest = Ki / 2 Therefore, the kinetic energy of the projectile at its highest point will be half of its initial kinetic energy, i.e., K.E-highest = Ki / 2 which can also be written as 0.5 Ki or as the question asked in terms of E, it will become 0.5 E, hence option A is correct.