Answer :
To solve this problem, we'll break it down into a few steps using the principles of kinematics.
1. Determine the time taken to reach the maximum height:
- We start by calculating the time taken for the stone to reach its maximum height where the velocity will be 0.
- The initial upward velocity ([tex]\(u\)[/tex]) is 45 m/s and the acceleration due to gravity ([tex]\(g\)[/tex]) is 10 m/s² (downward).
- At the maximum height, the final velocity ([tex]\(v\)[/tex]) is 0 m/s.
- Using the equation [tex]\(v = u - gt\)[/tex]:
[tex]\[ 0 = 45 - 10t \][/tex]
- Solving for [tex]\(t\)[/tex]:
[tex]\[ t = \frac{45}{10} = 4.5 \text{ seconds} \][/tex]
2. Calculate the maximum height reached above the tower:
- Next, we'll find the distance traveled upwards until the stone reaches this maximum height.
- Using the second equation of motion [tex]\( s = ut - \frac{1}{2}gt^2 \)[/tex]:
[tex]\[ s = 45 \cdot 4.5 - 0.5 \cdot 10 \cdot (4.5)^2 \][/tex]
- Substituting the values:
[tex]\[ s = 202.5 - 0.5 \cdot 10 \cdot 20.25 \][/tex]
[tex]\[ s = 202.5 - 101.25 = 101.25 \text{ meters} \][/tex]
3. Determine the time taken to fall back to the ground from the maximum height:
- The total time of flight is 10 seconds.
- Hence, time taken to descend from the maximum height back to the ground is the total time minus the ascent time:
[tex]\[ \text{Time to descend} = 10 - 4.5 = 5.5 \text{ seconds} \][/tex]
4. Calculate the distance fallen during the descent:
- To find the distance fallen from the maximum height back to the ground, we use the equation of motion for the descent:
[tex]\[ \text{Distance} = \frac{1}{2}gt^2 \][/tex]
- Substituting [tex]\( t = 5.5 \text{ seconds} \)[/tex] and [tex]\( g = 10 \text{ m/s}^2 \)[/tex]:
[tex]\[ \text{Distance} = 0.5 \cdot 10 \cdot (5.5)^2 = 5 \cdot 30.25 = 151.25 \text{ meters} \][/tex]
5. Calculate the total height of the tower:
- The total height of the tower is the sum of the distance during ascent and the distance during descent.
[tex]\[ \text{Total height} = 101.25 + 151.25 = 252.5 \text{ meters} \][/tex]
Thus, the height of the tower is [tex]\( 252.5 \)[/tex] meters.
1. Determine the time taken to reach the maximum height:
- We start by calculating the time taken for the stone to reach its maximum height where the velocity will be 0.
- The initial upward velocity ([tex]\(u\)[/tex]) is 45 m/s and the acceleration due to gravity ([tex]\(g\)[/tex]) is 10 m/s² (downward).
- At the maximum height, the final velocity ([tex]\(v\)[/tex]) is 0 m/s.
- Using the equation [tex]\(v = u - gt\)[/tex]:
[tex]\[ 0 = 45 - 10t \][/tex]
- Solving for [tex]\(t\)[/tex]:
[tex]\[ t = \frac{45}{10} = 4.5 \text{ seconds} \][/tex]
2. Calculate the maximum height reached above the tower:
- Next, we'll find the distance traveled upwards until the stone reaches this maximum height.
- Using the second equation of motion [tex]\( s = ut - \frac{1}{2}gt^2 \)[/tex]:
[tex]\[ s = 45 \cdot 4.5 - 0.5 \cdot 10 \cdot (4.5)^2 \][/tex]
- Substituting the values:
[tex]\[ s = 202.5 - 0.5 \cdot 10 \cdot 20.25 \][/tex]
[tex]\[ s = 202.5 - 101.25 = 101.25 \text{ meters} \][/tex]
3. Determine the time taken to fall back to the ground from the maximum height:
- The total time of flight is 10 seconds.
- Hence, time taken to descend from the maximum height back to the ground is the total time minus the ascent time:
[tex]\[ \text{Time to descend} = 10 - 4.5 = 5.5 \text{ seconds} \][/tex]
4. Calculate the distance fallen during the descent:
- To find the distance fallen from the maximum height back to the ground, we use the equation of motion for the descent:
[tex]\[ \text{Distance} = \frac{1}{2}gt^2 \][/tex]
- Substituting [tex]\( t = 5.5 \text{ seconds} \)[/tex] and [tex]\( g = 10 \text{ m/s}^2 \)[/tex]:
[tex]\[ \text{Distance} = 0.5 \cdot 10 \cdot (5.5)^2 = 5 \cdot 30.25 = 151.25 \text{ meters} \][/tex]
5. Calculate the total height of the tower:
- The total height of the tower is the sum of the distance during ascent and the distance during descent.
[tex]\[ \text{Total height} = 101.25 + 151.25 = 252.5 \text{ meters} \][/tex]
Thus, the height of the tower is [tex]\( 252.5 \)[/tex] meters.
