Answer
**a)** The initial speed \( u \) is approximately **35.2 m/s**.
**b)** The total time the particle is in motion is approximately **7 seconds**.
Solution
Certainly! Let's solve the problem step by step.
### **Problem Statement**
A particle is projected vertically upwards from point \( O \) with an initial speed \( u \) meters per second (\( \mathrm{m/s} \)). The greatest height reached by the particle is 62 meters above point \( O \).
We are to find:
a) The initial speed \( u \).
b) The total time the particle remains in motion.
---
### **Given Data**
- Maximum height (\( H \)) = 62 meters
- Acceleration due to gravity (\( g \)) ≈ \( 10 \, \mathrm{m/s}^2 \) (assuming \( g = 10 \, \mathrm{m/s}^2 \) for simplicity)
### **Part a: Finding the Initial Speed \( u \)**
When a particle is projected vertically upwards, the maximum height (\( H \)) it reaches can be determined using the kinematic equation:
\[
H = \frac{u^2}{2g}
\]
Rearranging the equation to solve for \( u \):
\[
u^2 = 2gH \\
u = \sqrt{2gH}
\]
Plugging in the given values:
\[
u = \sqrt{2 \times 10 \, \mathrm{m/s}^2 \times 62 \, \mathrm{m}} \\
u = \sqrt{1240} \\
u \approx 35.2 \, \mathrm{m/s}
\]
**Answer for Part a:**
The initial speed \( u \) is approximately **35.2 m/s**.
---
### **Part b: Finding the Total Time in Motion**
The total time the particle remains in motion consists of the time taken to reach the maximum height (\( t_{\text{up}} \)) and the time taken to descend back to point \( O \) (\( t_{\text{down}} \)).
1. **Time to Reach Maximum Height (\( t_{\text{up}} \)):**
\[
t_{\text{up}} = \frac{u}{g}
\]
2. **Time to Descend Back (\( t_{\text{down}} \)):**
The time to descend is equal to the time taken to ascend because of symmetry in projectile motion (assuming no air resistance).
\[
t_{\text{down}} = t_{\text{up}} = \frac{u}{g}
\]
3. **Total Time (\( T \)):**
\[
T = t_{\text{up}} + t_{\text{down}} = 2 \times \frac{u}{g}
\]
Substituting the known values:
\[
T = 2 \times \frac{35.2 \, \mathrm{m/s}}{10 \, \mathrm{m/s}^2} \\
T = 2 \times 3.52 \, \mathrm{s} \\
T = 7.04 \, \mathrm{s}
\]
Rounded to a suitable number of decimal places:
**Answer for Part b:**
The total time the particle is in motion is approximately **7 seconds**.
---
### **Summary of Answers**
**a)** The initial speed \( u \) is approximately **35.2 m/s**.
**b)** The total time the particle is in motion is approximately **7 seconds**.
Reviewed and approved by the UpStudy tutoring team
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