To address the problem of determining the maximum height attained by a body thrown vertically upwards and then returning to the ground in 4 seconds, follow these steps:
1. Understand Total Time:
- The total journey time for the body to go up and come back down is 4 seconds.
2. Calculate Time to Reach Maximum Height:
- The journey to the maximum height is half of the total journey. Therefore, the time to reach the maximum height is:
[tex]\[
\text{Time to maximum height} = \frac{\text{Total time}}{2} = \frac{4}{2} = 2 \text{ seconds}
\][/tex]
3. Use Gravity:
- The acceleration due to gravity (g) is approximately 9.8 m/s².
4. Determine Initial Velocity:
- At the peak of its flight, the velocity of the body becomes zero. Using the equation [tex]\(v = u - gt\)[/tex], where:
- [tex]\(v\)[/tex] is the final velocity (0 m/s at the peak point),
- [tex]\(u\)[/tex] is the initial velocity,
- [tex]\(g\)[/tex] is the acceleration due to gravity (9.8 m/s²),
- [tex]\(t\)[/tex] is the time to reach the peak (2 seconds).
- Rearranging for [tex]\(u\)[/tex]:
[tex]\[
0 = u - (9.8 \times 2)
\][/tex]
[tex]\[
u = 9.8 \times 2 = 19.6 \text{ m/s}
\][/tex]
5. Calculate Maximum Height:
- Using the kinematic formula for displacement:
[tex]\(h = ut - \frac{1}{2} gt^2\)[/tex]
- Where:
- [tex]\(h\)[/tex] is the maximum height,
- [tex]\(u\)[/tex] is the initial velocity (19.6 m/s),
- [tex]\(t\)[/tex] is the time to reach the maximum height (2 seconds),
- [tex]\(g\)[/tex] is the acceleration due to gravity (9.8 m/s²).
- Plugging in the values:
[tex]\[
h = (19.6 \times 2) - \frac{1}{2} \times 9.8 \times (2^2)
\][/tex]
[tex]\[
h = 39.2 - \frac{1}{2} \times 9.8 \times 4
\][/tex]
[tex]\[
h = 39.2 - 19.6
\][/tex]
[tex]\[
h = 19.6 \text{ meters}
\][/tex]
Therefore, the maximum height attained by the body is 19.6 meters.
