A 2.0-kg projectile is fired with initial velocity components `v_(o x)=30 m//s` and `v_(o y)=40 m//s` from a point on the earth's surface. Neglect any effects due to air resistance.
What is the kinetic energy of the projectile when it reaches the highest point in its trajectory ?

Options

A. zerojoules
B. 1600J
C. 900J
D. 2500J

Answer

To find the kinetic energy of the projectile when it reaches the highest point in its trajectory, we can follow these steps:

### Step 1: Understand the components of the initial velocity
The projectile is fired with two components of velocity:
- Horizontal component, \( v_{0x} = 30 \, \text{m/s} \)
- Vertical component, \( v_{0y} = 40 \, \text{m/s} \)

### Step 2: Determine the vertical velocity at the highest point
At the highest point of its trajectory, the vertical component of the velocity becomes zero (\( v_y = 0 \, \text{m/s} \)). The horizontal component remains unchanged throughout the motion (neglecting air resistance).

### Step 3: Calculate the horizontal velocity at the highest point
The horizontal velocity at the highest point is equal to the initial horizontal velocity:
- \( v_x = v_{0x} = 30 \, \text{m/s} \)

### Step 4: Use the kinetic energy formula
The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
where:
- \( m \) is the mass of the projectile,
- \( v \) is the velocity of the projectile at the highest point.

### Step 5: Substitute the values into the kinetic energy formula
Given:
- Mass, \( m = 2.0 \, \text{kg} \)
- Horizontal velocity at the highest point, \( v = 30 \, \text{m/s} \)

Now, substituting these values into the kinetic energy formula:
\[ KE = \frac{1}{2} \times 2.0 \, \text{kg} \times (30 \, \text{m/s})^2 \]

### Step 6: Calculate the kinetic energy
Calculating the square of the velocity:
\[ (30 \, \text{m/s})^2 = 900 \, \text{m}^2/\text{s}^2 \]
Now substituting this back into the equation:
\[ KE = \frac{1}{2} \times 2.0 \times 900 \]
\[ KE = 1.0 \times 900 = 900 \, \text{J} \]

### Final Answer
The kinetic energy of the projectile when it reaches the highest point in its trajectory is **900 Joules**.

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