A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively.
What is the magnitude of the projectile's velocity just before it strikes the ground?

Options

A. zero m/s
B. 30 m/s
C. 50 m/s
D. 9.8 m/s

Answer

To find the magnitude of the projectile's velocity just before it strikes the ground, we can follow these steps:

### Step 1: Identify the initial horizontal and vertical components of velocity.
The problem states that the initial horizontal component of velocity \( V_{x0} \) is 30 m/s and the initial vertical component of velocity \( V_{y0} \) is 40 m/s.

### Step 2: Understand the motion of the projectile.
In projectile motion, the horizontal component of velocity remains constant because there are no horizontal forces acting on the projectile (assuming air resistance is negligible). Thus, the horizontal component just before striking the ground will still be \( V_{x} = V_{x0} = 30 \, \text{m/s} \).

### Step 3: Determine the vertical component of velocity just before impact.
The vertical component of velocity changes due to the acceleration caused by gravity. The equation of motion for the vertical component is given by:
\[ V_{y}^2 = V_{y0}^2 + 2a s \]
where:
- \( V_{y} \) is the final vertical velocity,
- \( V_{y0} \) is the initial vertical velocity (40 m/s),
- \( a \) is the acceleration due to gravity (approximately \( -9.81 \, \text{m/s}^2 \)),
- \( s \) is the vertical displacement (which is the height from which the projectile is fired, but we can consider it as 0 when it hits the ground).

Since we want to find \( V_{y} \) just before it strikes the ground, we can use the following simplified equation:
\[ V_{y} = V_{y0} - g t \]
However, we can also use the energy conservation approach or simply recognize that the vertical component will be equal in magnitude but opposite in direction to the initial vertical component when it hits the ground. Therefore, we can conclude:
\[ V_{y} = V_{y0} = 40 \, \text{m/s} \]

### Step 4: Calculate the magnitude of the resultant velocity.
The magnitude of the resultant velocity \( V \) just before the projectile strikes the ground can be calculated using the Pythagorean theorem:
\[ V = \sqrt{V_{x}^2 + V_{y}^2} \]
Substituting the values we have:
\[ V = \sqrt{(30 \, \text{m/s})^2 + (40 \, \text{m/s})^2} \]
\[ V = \sqrt{900 + 1600} \]
\[ V = \sqrt{2500} \]
\[ V = 50 \, \text{m/s} \]

### Conclusion:
The magnitude of the projectile's velocity just before it strikes the ground is **50 m/s**.

---