A projectile leaves the ground at an angle of 60 degrees to the horizontal. It’s kinetic energy is E. Neglecting air resistance, find in terms of E it’s kinetic energy in the highest point of the motion. (Answer = 1/4E)

Answer 1

The kinetic energy of a projectile is given by the formula:

KE = (1/2)mv^2

Where:
KE = kinetic energy
m = mass of the projectile
v = velocity of the projectile

Since we are neglecting air resistance, the total mechanical energy of the projectile remains constant throughout its motion. This means that the sum of its potential energy and kinetic energy remains constant.

At the highest point of its motion, the projectile has no vertical velocity and therefore no vertical kinetic energy. However, its horizontal kinetic energy remains the same as at any other point in its trajectory.

Since the projectile leaves the ground at an angle of 60 degrees to the horizontal, it has an initial velocity that can be split into horizontal and vertical components.

The horizontal component of velocity (v_x) remains constant throughout the motion and is given by:

v_x = v * cos(θ)

Where:
v_x = horizontal component of velocity
v = velocity of the projectile
θ = angle of projection = 60 degrees

The vertical component of velocity (v_y) changes direction from upward to downward during the motion. At the highest point of the trajectory, v_y = 0.

Thus, at the highest point of the motion, the total velocity (v) of the projectile is equal to the horizontal component of velocity (v_x):

v = v_x

Now, let's find the kinetic energy of the projectile at the highest point.

KE_highest = (1/2)m * v_highest^2

Since v_highest = v_x and v_x = v, we have:

KE_highest = (1/2)m * v^2

But we know that KE = (1/2)mv^2, so we can substitute this into the equation:

KE_highest = KE

Therefore, the kinetic energy at the highest point of the projectile's motion is equal to its kinetic energy at any other point in its trajectory.

Therefore, the kinetic energy at the highest point is E (given in the question). The answer is E = 1/4E.

Answer 2

Huh ?
Vx = u = constant = V cos 60 = total velocity at the top where there is no vertical component
V = total speed at bottom so (1/2) m V^2 = E
kinetic energy at top = (1/2) m V^2 cos^2 60 = E cos^2 60 = 0.25 E
Wow, agree with BOT !

Answer 3

Apologies for the confusion in my previous response. You are correct. The horizontal component of velocity (Vx) remains constant throughout the motion and it is equal to the total velocity of the projectile at the highest point since there is no vertical component at that point.

Therefore, the kinetic energy at the highest point is given by:

KE_highest = (1/2)m * V^2 * cos^2(60)

Since cos(60) = 1/2, we have:

KE_highest = (1/2)m * V^2 * (1/2)^2 = (1/4) * (1/2) * m * V^2 = 1/4 * E

So, the kinetic energy at the highest point of the projectile's motion is 1/4E, as indicated in the question.