$$ v_{f}=20,93 \mathrm{~m} \cdot \mathrm{~s}^{-1} $$ $$ \Delta t= $$ Checkpoint 3 1. A ball is thrown upward from the top of a 30 m high building. The ball taker 24 s to reach its highest point. Ignore the effects of air resistance Calculate: a) the velocity with which the ball left the thrower's hand. b) the total time taken for the ball to reach the ground.

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### **Problem Statement**

A ball is thrown upward from the top of a **30 m high building**. The ball takes **24 seconds** to reach its highest point. Ignore the effects of air resistance. Calculate:

**a)** The velocity with which the ball left the thrower's hand.

**b)** The total time taken for the ball to reach the ground.

---

### **Given:**
- Height of the building, $$ y_0 = 30 \, \text{m} $$
- Time to reach the highest point, $$ t_{\text{up}} = 24 \, \text{s} $$
- Acceleration due to gravity, $$ g = 9.8 \, \text{m/s}^2 $$ (acting downward)

---

### **a) Calculating the Initial Velocity ($$ v_0 $$)**

When the ball reaches its highest point, its **final velocity ($$ v_f $$)** at that instant is **0 m/s**.

Using the kinematic equation:
$$
v_f = v_0 - g t_{\text{up}}
$$
Solving for $$ v_0 $$:
$$
v_0 = v_f + g t_{\text{up}} = 0 + (9.8 \, \text{m/s}^2)(24 \, \text{s}) = 235.2 \, \text{m/s}
$$
**Therefore, the initial velocity is **$$ \mathbf{235.2 \, \text{m/s} $$ upwards**.

---

### **b) Calculating the Total Time to Reach the Ground**

The motion of the ball can be divided into two phases:

1. **Ascending to the Highest Point**
   - **Time:** $$ t_{\text{up}} = 24 \, \text{s} $$
   - **Displacement during ascent ($$ s_{\text{up}} $$):**
     $$
     s_{\text{up}} = v_0 t_{\text{up}} - \frac{1}{2} g t_{\text{up}}^2 = (235.2 \, \text{m/s})(24 \, \text{s}) - \frac{1}{2}(9.8 \, \text{m/s}^2)(24)^2
     $$
     $$
     s_{\text{up}} = 5644.8 \, \text{m} - 2822.4 \, \text{m} = 2822.4 \, \text{m}
     $$
   - **Maximum Height ($$ y_{\text{max}} $$):**
     $$
     y_{\text{max}} = y_0 + s_{\text{up}} = 30 \, \text{m} + 2822.4 \, \text{m} = 2852.4 \, \text{m}
     $$

2. **Descending from the Highest Point to the Ground**
   - **Displacement during descent ($$ s_{\text{down}} $$):** $$ 2852.4 \, \text{m} $$ (since it falls back to the ground level)
   - **Time to descend ($$ t_{\text{down}} $$)** can be found using:
     $$
     s_{\text{down}} = \frac{1}{2} g t_{\text{down}}^2
     $$
     Solving for $$ t_{\text{down}} $$:
     $$
     t_{\text{down}} = \sqrt{\frac{2 s_{\text{down}}}{g}} = \sqrt{\frac{2 \times 2852.4 \, \text{m}}{9.8 \, \text{m/s}^2}} \approx \sqrt{\frac{5704.8}{9.8}} \approx \sqrt{582.0} \approx 24.13 \, \text{s}
     $$

**Total Time ($$ t_{\text{total}} $$)** is the sum of ascending and descending times:
$$
t_{\text{total}} = t_{\text{up}} + t_{\text{down}} = 24 \, \text{s} + 24.13 \, \text{s} \approx 48.13 \, \text{s}
$$

**Therefore, the total time taken for the ball to reach the ground is approximately **$$ \mathbf{48.1 \, \text{seconds}} $$**.

---

### **Summary**
- **a)** Initial Velocity: **235.2 m/s upwards**
- **b)** Total Time to Reach the Ground: **Approximately 48.1 seconds**

---

**Note:** The computed initial velocity of **235.2 m/s** is exceptionally high for a typical athletic throw, suggesting that either the time to reach the highest point might have been misstated or additional factors are at play. In real-world scenarios, such a velocity is impractical for a thrown object and would require specialized equipment or conditions.
**a)** The ball was thrown upwards with an initial velocity of **235.2 m/s**. **b)** The total time taken for the ball to reach the ground is approximately **48.1 seconds**.

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