a projectile is fired with initial speed of
1
5
0
m
/
s and an angle of elevation
4
5
degrees from a position m above ground level. Where does the projectile hit the ground and with what speed

Question

a projectile is fired with initial speed of 
1
5
0
 ﻿m
/
s and an angle of elevation 
4
5
 ﻿degrees from a position m above ground level. Where does the projectile hit the ground and with what speed

give the calculation step by step as like as full answer in 2 steps and explanation block in every steps. 
And final answer summary
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VIPSolutions · Accepted Answer

**Definition of Parameters**  
Let  
- $ v_0 = 150 $ m/s (initial speed)  
- $ \theta = 45^\circ $ (angle of elevation)  
- $ h = m $ (initial height above ground; left as $ m $ since no value provided)  
- $ g = 9.8 $ m/s² (acceleration due to gravity)

The projectile is fired from a height $ h = m $ above ground. The task is to find:  
1. The horizontal distance where it hits the ground (range, $ R $).  
2. The speed on impact with the ground.

---

### **Step 1: Find the Time of Flight**

**Calculation:**  
- The initial velocity components:  
  - $ v_{0x} = v_0 \cos \theta = 150 \cos 45^\circ = 150 \times \frac{1}{\sqrt{2}} \approx 106.07 $ m/s  
  - $ v_{0y} = v_0 \sin \theta = 150 \sin 45^\circ = 150 \times \frac{1}{\sqrt{2}} \approx 106.07 $ m/s

- The vertical position as a function of time ($ y(t) $), taking upward as positive:  
  - $ y(t) = h + v_{0y} t - \frac{1}{2} g t^2 $

- The projectile hits the ground when $ y(t) = 0 $:  
  $$
  0 = h + v_{0y} t - \frac{1}{2} g t^2 
  $$
  $$
  \frac{1}{2}g t^2 - v_{0y} t - h = 0
  $$
  This is a quadratic in $ t $:
  $$
  t = \frac{v_{0y} + \sqrt{v_{0y}^2 + 2gh}}{g}
  $$
  (The negative root corresponds to a time before launch.)

Substitute values:
  $$
  t = \frac{106.07 + \sqrt{(106.07)^2 + 2 \times 9.8 \times h}}{9.8}
  $$
  $$
  t = \frac{106.07 + \sqrt{11250 + 19.6h}}{9.8}
  $$

**Explanation:**  
The time of flight is found by solving for when the vertical position reaches ground level. This requires solving a quadratic equation for time, with the positive root giving the physically meaningful answer.

---

### **Step 2: Find the Horizontal Range and Impact Speed**

**Calculation:**  
**A. Horizontal Range ($ R $):**  
$$
R = v_{0x} \cdot t = 106.07 \cdot t
$$
Substitute $ t $ from above:
$$
R = 106.07 \times \frac{106.07 + \sqrt{11250 + 19.6h}}{9.8}
$$

**B. Speed on Impact:**  
At impact, vertical velocity:
$$
v_y = v_{0y} - g t
$$
Horizontal velocity remains constant:
$$
v_x = v_{0x}
$$
Speed on impact:
$$
v_{\text{impact}} = \sqrt{v_x^2 + v_y^2}
$$

Substitute $ v_{0x} = 106.07 $ m/s, $ v_{0y} = 106.07 $ m/s, and $ t $ from above.

$$
v_y = 106.07 - 9.8 t
$$
$$
v_{\text{impact}} = \sqrt{106.07^2 + (106.07 - 9.8 t)^2}
$$

**Explanation:**  
The horizontal range is determined by multiplying the horizontal velocity by the time of flight. The speed on impact is found by combining the unchanged horizontal velocity with the vertical velocity at the moment of hitting the ground, using the Pythagorean theorem.

---

### **Final Answer Summary**

- **Horizontal Range ($ R $)**:  
  $$
  R = 106.07 \times \frac{106.07 + \sqrt{11250 + 19.6h}}{9.8} \;\; \text{meters}
  $$
  (where $ h $ is the initial height above ground in meters)

- **Speed on Impact ($ v_{\text{impact}} $)**:  
  $$
  v_{\text{impact}} = \sqrt{106.07^2 + (106.07 - 9.8 t)^2} \;\; \text{m/s}
  $$
  with  
  $$
  t = \frac{106.07 + \sqrt{11250 + 19.6h}}{9.8}
  $$

**Note:**  
For a specific value of $ h $, substitute and compute numerically. All steps are shown with clear logical progression and explanation for each step.

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