A projectile is fired with an initial speed of 38.0 m/sm/s at an angle of 42.7 ∘∘ above the horizontal on a long flat firing range. Determine the speed of the projectile 1.50 ss after firing.

To find the speed of the projectile 1.50 seconds after firing, we need to break down the initial velocity into horizontal and vertical components, then determine the velocity at that time using the kinematic equations.

### Step 1: Break the initial velocity into components

Given:
- Initial speed \( v_0 = 38.0 \, \text{m/s} \)
- Angle \( \theta = 42.7^\circ \)
- Time \( t = 1.50 \, \text{s} \)
- Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)

We need to resolve the initial velocity into horizontal and vertical components:

\[
v_{0x} = v_0 \cdot \cos(\theta) = 38.0 \cdot \cos(42.7^\circ)
\]
\[
v_{0y} = v_0 \cdot \sin(\theta) = 38.0 \cdot \sin(42.7^\circ)
\]

### Step 2: Calculate the horizontal and vertical components

Using trigonometric values:
\[
\cos(42.7^\circ) \approx 0.736 \quad \text{and} \quad \sin(42.7^\circ) \approx 0.677
\]

Now, calculate the components:
\[
v_{0x} = 38.0 \times 0.736 \approx 27.97 \, \text{m/s}
\]
\[
v_{0y} = 38.0 \times 0.677 \approx 25.73 \, \text{m/s}
\]

### Step 3: Determine the velocity components after 1.50 seconds

#### Horizontal component (\( v_x \)):
Since there is no horizontal acceleration (ignoring air resistance), the horizontal velocity remains constant:
\[
v_x = v_{0x} = 27.97 \, \text{m/s}
\]

#### Vertical component (\( v_y \)):
The vertical velocity changes due to gravity. The formula to calculate the vertical velocity after time \( t \) is:

\[
v_y = v_{0y} - g \cdot t
\]

Substitute the values:
\[
v_y = 25.73 \, \text{m/s} - 9.8 \, \text{m/s}^2 \times 1.50 \, \text{s}
\]
\[
v_y = 25.73 \, \text{m/s} - 14.7 \, \text{m/s}
\]
\[
v_y \approx 11.03 \, \text{m/s}
\]

### Step 4: Calculate the magnitude of the velocity (speed)

Now that we have both the horizontal and vertical components of the velocity at \( t = 1.50 \, \text{s} \), we can calculate the resultant speed using the Pythagorean theorem:

\[
v = \sqrt{v_x^2 + v_y^2}
\]
\[
v = \sqrt{(27.97)^2 + (11.03)^2}
\]
\[
v = \sqrt{782.72 + 121.66}
\]
\[
v = \sqrt{904.38}
\]
\[
v \approx 30.07 \, \text{m/s}
\]

### Final Answer:

The speed of the projectile 1.50 seconds after firing is approximately **30.1 m/s** (rounded to three significant figures).

Would you like more details or have any questions? Here are five related questions for further practice:

1. What is the horizontal displacement of the projectile after 1.50 seconds?
2. What is the maximum height reached by the projectile?
3. How long does the projectile stay in the air before hitting the ground?
4. What is the total horizontal distance traveled by the projectile (its range)?
5. How fast is the projectile moving at its maximum height?

**Tip:** When calculating projectile motion, always separate the motion into horizontal and vertical components and treat them independently.

A projectile is fired with an initial speed of 38.0 m/s at an angle of 42.7° above the horizontal on a long flat firing range. Determine the speed of the projectile 1.50 seconds after firing.

Determine the speed of a projectile fired at 38.0 m/s with an angle of 42.7° after 1.50 seconds. Learn how to break down velocity components and apply the Pythagorean theorem to find the resultant speed. Suitable for students in Grades 9-12.

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