If the kinetic energy of an oblique projectile at its maximum height is half of its initial kinetic energy then the angle of throw with the vertical is

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if the kinetic energy of an oblique projectile at its maximum height is half of its initial kinetic energy then the angle of throw with the vertical is

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If the kinetic energy of an oblique projectile at its maximum height is half of its initial kinetic energy, then the angle of throw with the vertical can be determined.

Let’s analyze the situation using the principles of projectile motion.

When a projectile is launched at an angle with the vertical, it follows a curved path. At the maximum height of its trajectory, the projectile momentarily comes to a stop before falling back down due to gravity.

Given that the kinetic energy at the maximum height is half of the initial kinetic energy, we can use the conservation of mechanical energy to find the angle of throw.

The mechanical energy of a projectile is the sum of its potential energy (PE) and kinetic energy (KE). At the maximum height, the kinetic energy is zero, so all the initial kinetic energy is converted into potential energy.

The equation for mechanical energy is:

ME = PE + KE

Since the kinetic energy at the maximum height is half of the initial kinetic energy, we can write:

ME = 0.5 * KE_initial

At the maximum height, the potential energy is at its maximum value, which is equal to the initial kinetic energy. Therefore, we can also write:

ME = PE_max

Setting these two equations equal to each other, we have:

0.5 * KE_initial = PE_max

Now, let’s look at the components of potential energy and kinetic energy in terms of the launch angle theta.

PE = m * g * h
KE = 0.5 * m * v^2

Where:
m is the mass of the projectile,
g is the acceleration due to gravity,
h is the maximum height,
v is the initial velocity of the projectile.

The initial velocity can be split into vertical and horizontal components:

v_initial = v_x + v_y

Using trigonometry, we can express the vertical and horizontal components as:

v_x = v_initial * cos(theta)
v_y = v_initial * sin(theta)

The maximum height can be found using the equation for vertical motion:

v_y^2 = (v_initial * sin(theta))^2 - 2 * g * h_max

Since the projectile comes to a stop at the maximum height, v_y is equal to zero:

0 = (v_initial * sin(theta))^2 - 2 * g * h_max

From this equation, we can solve for h_max:

h_max = (v_initial^2 * sin^2(theta)) / (2 * g)

Now we can substitute the expressions for PE_max and KE_initial in terms of h_max and v_initial:

0.5 * KE_initial = PE_max
0.5 * (0.5 * m * v_initial^2) = m * g * h_max

Simplifying the equation, we have:

0.25 * v_initial^2 = 2 * g * h_max

Substituting the expression for h_max, we get:

0.25 * v_initial^2 = 2 * g * ((v_initial^2 * sin^2(theta)) / (2 * g))

Canceling out terms and simplifying, we obtain:

0.25 = sin^2(theta)

Taking the square root of both sides, we have:

0.5 = sin(theta)

Thus, the angle of throw with the vertical is 30 degrees.

So, if the kinetic energy of an oblique projectile at its maximum height is half of its initial kinetic energy, then the angle of throw with the vertical is 30 degrees.