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.
How much work was done in firing the projectile ?

Options

A. 900J
B. 2500J
C. 1600J
D. 4900J

Answer

To find the work done in firing a projectile, we can use the following steps:

### Step 1: Understand the Problem
We have a projectile of mass \( m = 2.0 \, \text{kg} \) that is fired with initial velocity components \( v_{0x} = 30 \, \text{m/s} \) and \( v_{0y} = 40 \, \text{m/s} \). We need to calculate the work done in firing this projectile until it reaches its highest point.

### Step 2: Identify the Forces Acting on the Projectile
The only force acting on the projectile after it is fired is the gravitational force, which acts downward. The gravitational force \( F_g \) can be calculated as:
\[ F_g = m \cdot g \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 3: Calculate the Gravitational Force
Substituting the values:
\[ F_g = 2.0 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 20 \, \text{N} \]

### Step 4: Determine the Maximum Height
To find the maximum height \( H \) reached by the projectile, we can use the kinematic equation:
\[ v_y^2 = u_y^2 + 2a_y s \]
At the maximum height, the final vertical velocity \( v_y = 0 \). Thus, the equation simplifies to:
\[ 0 = (40 \, \text{m/s})^2 - 2 \cdot 10 \, \text{m/s}^2 \cdot H \]
Rearranging gives:
\[ H = \frac{(40 \, \text{m/s})^2}{2 \cdot 10 \, \text{m/s}^2} = \frac{1600}{20} = 80 \, \text{m} \]

### Step 5: Calculate the Work Done
The work done \( W \) against gravity to raise the projectile to its maximum height is given by:
\[ W = F_g \cdot H \]
Substituting the values we have:
\[ W = 20 \, \text{N} \cdot 80 \, \text{m} = 1600 \, \text{J} \]

### Conclusion
The work done in firing the projectile is \( 1600 \, \text{J} \).

---