Igmore the effects of air resistance. 1. A body is projected vertically upward from the roof of a building at $$ 40 \mathrm{~m} \cdot \mathrm{~s}^{-1} $$. It reaches the ground with a speed of $$ 60 \mathrm{~m} \cdot \mathrm{~s}^{-1} $$. Calculate: a) the height of the building. b) the total time of flight.

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UpStudy AI · Accepted Answer

Sure, let's solve the problem step by step by ignoring air resistance. We'll use the standard equations of motion under constant acceleration due to gravity.

### Given:
- **Initial velocity upwards ($$ u $$)**: $$ 40 \, \text{m/s} $$
- **Final velocity when reaching the ground ($$ v $$)**: $$ 60 \, \text{m/s} $$ downward
- **Acceleration due to gravity ($$ g $$)**: $$ 10 \, \text{m/s}^2 $$ (approximated for simplicity)

**Note:** We'll take upward as the positive direction. Therefore, the final velocity $$ v $$ will be negative since it's downward.

### Part (a): Height of the Building

We can use the **kinematic equation** that relates velocity and displacement:

$$
v^2 = u^2 + 2a s
$$

Where:
- $$ v = -60 \, \text{m/s} $$ (negative because it's downward)
- $$ u = 40 \, \text{m/s} $$
- $$ a = -g = -10 \, \text{m/s}^2 $$ (acceleration is downward)
- $$ s = -h $$ (displacement is downward, hence negative)

Plugging in the values:

$$
(-60)^2 = (40)^2 + 2(-10)(-h)
$$
$$
3600 = 1600 + 20h
$$
$$
3600 - 1600 = 20h
$$
$$
2000 = 20h
$$
$$
h = \frac{2000}{20} = 100 \, \text{meters}
$$

**Height of the building ($$ h $$) is 100 meters.**

### Part (b): Total Time of Flight

The total time of flight can be divided into two phases:
1. **Going Up:** Time to reach the maximum height.
2. **Coming Down:** Time to descend from the maximum height to the ground.

**1. Time to Reach Maximum Height ($$ t_1 $$):**

At maximum height, the final velocity is 0.

$$
v = u + at
$$
$$
0 = 40 - 10t_1
$$
$$
t_1 = \frac{40}{10} = 4 \, \text{seconds}
$$

**2. Time to Descend from Maximum Height to Ground ($$ t_2 $$):**

The total displacement from the maximum height to the ground is $$ h + h_1 $$, where $$ h_1 $$ is the additional height gained.

First, calculate $$ h_1 $$:

$$
h_1 = ut_1 + \frac{1}{2} a t_1^2 = 40 \times 4 + \frac{1}{2} \times (-10) \times 4^2 = 160 - 80 = 80 \, \text{meters}
$$

Total height from the ground:

$$
H = h + h_1 = 100 + 80 = 180 \, \text{meters}
$$

Now, calculate $$ t_2 $$ using the equation:

$$
s = \frac{1}{2} g t_2^2
$$
$$
180 = \frac{1}{2} \times 10 \times t_2^2
$$
$$
180 = 5 t_2^2
$$
$$
t_2^2 = \frac{180}{5} = 36
$$
$$
t_2 = \sqrt{36} = 6 \, \text{seconds}
$$

**Total Time of Flight ($$ T $$)**:

$$
T = t_1 + t_2 = 4 + 6 = 10 \, \text{seconds}
$$

### **Summary of Answers:**
- **a) Height of the building:** $$ 100 \, \text{meters} $$
- **b) Total time of flight:** $$ 10 \, \text{seconds} $$
The height of the building is 100 meters, and the total time of flight is 10 seconds.

Igmore the effects of air resistance.

A body is projected vertically upward from the roof of a building at .
It reaches the ground with a speed of . Calculate:
a) the height of the building.
b) the total time of flight.

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a) Height of the Building

b) Total Time of Flight

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Igmore the effects of air resistance. A body is projected vertically upward from the roof of a building at . It reaches the ground with a speed of . Calculate: a) the height of the building. b) the total time of flight.