average Displacement : 5.7meter time : 1.6 s (calculate : height of bullet. Initial velocity. Angle of firing.)

y-direction:

S= 5.7m

U=

V=

A= -9.81 ms^-2

T= 1.6s

X-direction

S=5.7m

U=

V= 0ms^-2

A=

T=1.6s

the data is taken by a catapult

To calculate the **initial velocity**, **height**, and **angle of firing** for the bullet launched from the catapult, we can break down the motion into two components: **vertical (y-direction)** and **horizontal (x-direction)**.

### Horizontal Motion (x-direction):
- **Displacement (Sₓ)**: 5.7 m
- **Time (T)**: 1.6 s
- **Initial velocity in the x-direction (Uₓ)**: To be determined
- **Final velocity in the x-direction (Vₓ)**: Constant (since there's no acceleration, we can assume it stays the same).
- **Acceleration (Aₓ)**: 0 m/s² (since we assume no air resistance)

Using the equation of motion for constant velocity in the x-direction:

\[
Sₓ = Uₓ \cdot T
\]
\[
Uₓ = \frac{Sₓ}{T}
\]
Substitute the known values:
\[
Uₓ = \frac{5.7 \, \text{m}}{1.6 \, \text{s}} = 3.56 \, \text{m/s}
\]

Thus, the horizontal component of the velocity is:
\[
Uₓ = 3.56 \, \text{m/s}
\]

---

### Vertical Motion (y-direction):
- **Displacement (Sᵧ)**: 5.7 m
- **Acceleration (Aᵧ)**: -9.81 m/s² (gravitational acceleration)
- **Time (T)**: 1.6 s
- **Initial velocity in the y-direction (Uᵧ)**: To be determined
- **Final velocity in the y-direction (Vᵧ)**: To be determined

Using the second equation of motion for vertical displacement:

\[
Sᵧ = Uᵧ \cdot T + \frac{1}{2} Aᵧ \cdot T^2
\]

Substitute the known values:
\[
5.7 = Uᵧ \cdot 1.6 + \frac{1}{2} \cdot (-9.81) \cdot (1.6)^2
\]
Simplifying:
\[
5.7 = 1.6 Uᵧ - 12.57
\]
\[
1.6 Uᵧ = 18.27
\]
\[
Uᵧ = \frac{18.27}{1.6} = 11.42 \, \text{m/s}
\]

Thus, the vertical component of the velocity is:
\[
Uᵧ = 11.42 \, \text{m/s}
\]

---

### Initial Velocity (Magnitude):

The initial velocity (U) is the vector sum of the horizontal and vertical components:
\[
U = \sqrt{Uₓ^2 + Uᵧ^2}
\]
\[
U = \sqrt{(3.56)^2 + (11.42)^2} = \sqrt{12.67 + 130.43} = \sqrt{143.1}
\]
\[
U = 11.96 \, \text{m/s}
\]

---

### Angle of Firing (θ):

The angle (θ) of the firing can be calculated using the tangent formula:
\[
\tan(\theta) = \frac{Uᵧ}{Uₓ}
\]
\[
\theta = \tan^{-1} \left( \frac{11.42}{3.56} \right)
\]
\[
\theta = \tan^{-1}(3.21)
\]
\[
\theta \approx 72.4^\circ
\]

---

### Summary of Results:
- **Initial Velocity (U)**: 11.96 m/s
- **Angle of Firing (θ)**: 72.4°
- **Height of bullet (Sᵧ)**: 5.7 m

---

Would you like more details or further clarifications? Here are some related questions you might find interesting:

1. What is the significance of the angle of projection in projectile motion?
2. How would the result change if air resistance was considered?
3. How does time of flight affect the horizontal and vertical distances covered?
4. How can you determine the maximum height reached by the projectile?
5. How do horizontal and vertical motions relate to the total distance traveled?

**Tip:** In projectile motion, horizontal and vertical components are independent of each other, except for the shared time of flight!

average Displacement: 5.7 meter, time: 1.6 s (calculate: height of bullet, initial velocity, angle of firing). y-direction: S= 5.7m, U=, V=, A= -9.81 ms^-2, T= 1.6s. X-direction: S=5.7m, U=, V= 0ms^-2, A=, T=1.6s. The data is taken by a catapult.

Learn how to calculate the initial velocity, height, and angle of firing of a bullet launched by a catapult with a displacement of 5.7 meters and a time of 1.6 seconds. This problem involves key concepts in projectile motion and kinematics, suitable for students in Grades 10-12.

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