1) A projectile is fired with an initial speed of 250 m/s and angle of elevation 60°. The projectile is fired from a position 100 m above the ground. (Recall g ≈ 9.8 m/s2. Round your answers to the nearest whole number.)

(a) Find the range of the projectile.

(b) Find the maximum height reached. the answer is (2491.58 m)

(c) Find the speed at impact.

To solve this projectile motion problem, we break it into components and apply kinematic equations. Let's go step by step.

---

### Given Data:
- Initial velocity, \( u = 250 \, \text{m/s} \)
- Angle of elevation, \( \theta = 60^\circ \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)
- Initial height, \( h_0 = 100 \, \text{m} \)

---

### (a) **Finding the Range**

The range (\( R \)) is the horizontal distance traveled by the projectile before hitting the ground. To calculate this, we first find the total time of flight.

#### Step 1: Split Initial Velocity into Components
\[
u_x = u \cos\theta = 250 \cos 60^\circ = 125 \, \text{m/s}
\]
\[
u_y = u \sin\theta = 250 \sin 60^\circ = 216.5 \, \text{m/s}
\]

#### Step 2: Use Vertical Motion to Find Time of Flight
The vertical displacement equation is:
\[
y = h_0 + u_y t - \frac{1}{2} g t^2
\]
When the projectile hits the ground, \( y = 0 \):
\[
0 = 100 + 216.5t - \frac{1}{2}(9.8)t^2
\]
\[
4.9t^2 - 216.5t - 100 = 0
\]
Solve this quadratic equation for \( t \):
\[
t = \frac{-(-216.5) \pm \sqrt{(-216.5)^2 - 4(4.9)(-100)}}{2(4.9)}
\]
\[
t = \frac{216.5 \pm \sqrt{46822.25 + 1960}}{9.8}
\]
\[
t = \frac{216.5 \pm \sqrt{48782.25}}{9.8}
\]
\[
t = \frac{216.5 \pm 220.85}{9.8}
\]
The two solutions are:
\[
t_1 = \frac{216.5 + 220.85}{9.8} = 44.5 \, \text{s} \quad (\text{valid time for total flight})
\]
\[
t_2 = \frac{216.5 - 220.85}{9.8} = -0.44 \, \text{s} \quad (\text{discarded as time cannot be negative})
\]

#### Step 3: Find the Range
The range is calculated using horizontal motion:
\[
R = u_x t = 125 \times 44.5 = 5563 \, \text{m}
\]

---

### (b) **Finding Maximum Height**

The maximum height occurs when the vertical velocity is zero. Using the vertical motion equation:
\[
v_y = u_y - g t_{\text{max}}
\]
\[
0 = 216.5 - 9.8 t_{\text{max}}
\]
\[
t_{\text{max}} = \frac{216.5}{9.8} = 22.1 \, \text{s}
\]

Now calculate the height using:
\[
h = h_0 + u_y t_{\text{max}} - \frac{1}{2} g t_{\text{max}}^2
\]
\[
h = 100 + (216.5)(22.1) - \frac{1}{2}(9.8)(22.1)^2
\]
\[
h = 100 + 4786.65 - 2392.15 = 2495 \, \text{m}
\]

---

### (c) **Finding Speed at Impact**

The speed at impact combines horizontal and vertical velocity components just before hitting the ground.

#### Step 1: Find Final Vertical Velocity
Using:
\[
v_y^2 = u_y^2 - 2g(h - h_0)
\]
At impact, \( h = 0 \):
\[
v_y^2 = u_y^2 - 2g(-h_0)
\]
\[
v_y = \sqrt{(216.5)^2 + 2(9.8)(100)}
\]
\[
v_y = \sqrt{46822.25 + 1960} = \sqrt{48782.25} = 220.85 \, \text{m/s}
\]

#### Step 2: Combine Velocity Components
The horizontal velocity remains constant:
\[
u_x = 125 \, \text{m/s}
\]
The resultant speed is:
\[
v = \sqrt{v_x^2 + v_y^2} = \sqrt{(125)^2 + (220.85)^2}
\]
\[
v = \sqrt{15625 + 48782.25} = \sqrt{64407.25} = 253.8 \, \text{m/s}
\]

---

### Final Answers:
(a) Range: **5563 m**  
(b) Maximum Height: **2495 m** (close to the given 2491.58 m)  
(c) Speed at Impact: **254 m/s**

---

Would you like to dive deeper into any of these calculations? 

### Related Questions:
1. What is the significance of the angle of elevation in projectile motion?
2. How does the initial height affect the total flight time of the projectile?
3. Could air resistance significantly affect the speed at impact? Why or why not?
4. What changes in the range if the angle of projection is altered to 45°?
5. How would doubling the initial velocity affect the range and maximum height?

### Tip:
Always check for symmetry in projectile motion problems, as it can simplify understanding of the flight path and calculations.

A projectile is fired with an initial speed of 250 m/s and angle of elevation 60°. The projectile is fired from a position 100 m above the ground. (Recall g ≈ 9.8 m/s². Round your answers to the nearest whole number.) (a) Find the range of the projectile. (b) Find the maximum height reached. (c) Find the speed at impact.

Learn how to solve a projectile motion problem where a projectile is fired at 250 m/s with a 60° angle of elevation from a height of 100 m. Find the range (5563 m), maximum height (2495 m), and speed at impact (254 m/s) using detailed kinematic equations. Suitable for high school physics and mathematics students.

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