Question

A projectile is thrown at an angle 60° above the horizontal and with kinetic energy. The kinetic energy of the projectile at the highest point of its trajectory will be:

• A. 10 J
• B. 40 J
• C. 20 J
• D. 20√2 J
• E. 20√3 J

Answer 1

The Correct Option is A

Given:

• Projectile launched at \( \theta = 60^\circ \) with initial kinetic energy \( K_0 = 40 \, \text{J} \).

Step 1: Resolve Initial Velocity

At the highest point, the vertical velocity component is zero, and only the horizontal component remains:

\[ v_x = v_0 \cos \theta \]

Step 2: Relate Initial KE to Final KE

Initial KE: \( K_0 = \frac{1}{2} m v_0^2 = 40 \, \text{J} \).

KE at highest point: \( K_{\text{top}} = \frac{1}{2} m v_x^2 = \frac{1}{2} m (v_0 \cos 60^\circ)^2 \).

Since \( \cos 60^\circ = 0.5 \):

\[ K_{\text{top}} = K_0 \cos^2 60^\circ = 40 \times (0.5)^2 = 10 \, \text{J} \]

Conclusion:

The kinetic energy at the highest point is 10 J.

Answer: \(\boxed{A}\)

Answer 2

1. Analyze the projectile motion:

At the highest point of its trajectory, the projectile's vertical velocity component is zero. However, the horizontal velocity component remains constant throughout the motion (assuming no air resistance).

2. Relate kinetic energy to velocity components:

The initial kinetic energy (KE_i) is given by:

\[KE_i = \frac{1}{2}mv_i^2 = 40 \, J\]

where \(v_i\) is the initial velocity.

The velocity can be resolved into horizontal (\(v_{ix}\)) and vertical (\(v_{iy}\)) components:

\[v_{ix} = v_i \cos(60^\circ) = \frac{1}{2}v_i\]

\[v_{iy} = v_i \sin(60^\circ) = \frac{\sqrt{3}}{2}v_i\]

3. Calculate the kinetic energy at the highest point:

At the highest point, the vertical velocity is zero (\(v_y = 0\)), and the horizontal velocity remains unchanged (\(v_x = v_{ix} = \frac{1}{2}v_i\)). The kinetic energy at the highest point (KE_h) is:

\[KE_h = \frac{1}{2}mv_x^2 = \frac{1}{2}m(\frac{1}{2}v_i)^2 = \frac{1}{4}(\frac{1}{2}mv_i^2) = \frac{1}{4}KE_i\]

4. Substitute the initial kinetic energy:

\[KE_h = \frac{1}{4}(40 \, J) = 10 \, J\]