Answer :
Sure, let's go through the problem step-by-step to find the ratio of times it takes for a particle to fall equal heights [tex]\( h \)[/tex] from rest when released from a tower of height [tex]\( 3h \)[/tex].
### Step 1: Understand the Problem
A particle is released from rest from the top of a tower of height [tex]\( 3h \)[/tex]. We need to find the ratio of the times taken to fall each height [tex]\( h \)[/tex].
### Step 2: Equations of Motion
We will use the basic equation of motion for freely falling objects:
[tex]\[ h = \frac{1}{2} g t^2 \][/tex]
where:
- [tex]\( h \)[/tex] is the height,
- [tex]\( g \)[/tex] is the acceleration due to gravity ([tex]\( \approx 9.8 \, \text{m/s}^2 \)[/tex]),
- [tex]\( t \)[/tex] is the time taken to fall that height.
### Step 3: Time to Fall Each Height [tex]\( h \)[/tex]
We need to find the times [tex]\( t_1 \)[/tex], [tex]\( t_2 \)[/tex], and [tex]\( t_3 \)[/tex] to fall each height [tex]\( h \)[/tex].
#### Time [tex]\( t_1 \)[/tex] to fall the first height [tex]\( h \)[/tex]:
From [tex]\( 3h \)[/tex] to [tex]\( 2h \)[/tex], the particle falls the first height [tex]\( h \)[/tex].
Using the motion equation:
[tex]\[ h = \frac{1}{2} g t_1^2 \][/tex]
Solving for [tex]\( t_1 \)[/tex]:
[tex]\[ t_1 = \sqrt{\frac{2h}{g}} \][/tex]
#### Time [tex]\( t_2 \)[/tex] to fall the second height [tex]\( h \)[/tex]:
From [tex]\( 2h \)[/tex] to [tex]\( h \)[/tex], the particle falls the second height [tex]\( h \)[/tex]. This time is the difference between the time to fall [tex]\( 2h \)[/tex] and the time to fall [tex]\( h \)[/tex].
[tex]\[ 2h = \frac{1}{2} g t_{2, \text{total}}^2 \][/tex]
Solving for [tex]\( t_{2, \text{total}} \)[/tex]:
[tex]\[ t_{2, \text{total}} = \sqrt{\frac{4h}{g}} \][/tex]
The time [tex]\( t_2 \)[/tex] to fall the second height [tex]\( h \)[/tex] is:
[tex]\[ t_2 = t_{2, \text{total}} - t_1 \][/tex]
[tex]\[ t_2 = \sqrt{\frac{4h}{g}} - \sqrt{\frac{2h}{g}} \][/tex]
#### Time [tex]\( t_3 \)[/tex] to fall the third height [tex]\( h \)[/tex]:
From [tex]\( h \)[/tex] to [tex]\( 0 \)[/tex], the particle falls the third height [tex]\( h \)[/tex]. This time is the difference between the time to fall [tex]\( 3h \)[/tex] and the time to fall [tex]\( 2h \)[/tex].
[tex]\[ 3h = \frac{1}{2} g t_{3, \text{total}}^2 \][/tex]
Solving for [tex]\( t_{3, \text{total}} \)[/tex]:
[tex]\[ t_{3, \text{total}} = \sqrt{\frac{6h}{g}} \][/tex]
The time [tex]\( t_3 \)[/tex] to fall the third height [tex]\( h \)[/tex] is:
[tex]\[ t_3 = t_{3, \text{total}} - t_{2, \text{total}} \][/tex]
[tex]\[ t_3 = \sqrt{\frac{6h}{g}} - \sqrt{\frac{4h}{g}} \][/tex]
### Step 4: Find the Ratio [tex]\( t_1 : t_2 : t_3 \)[/tex]
Substituting the values we calculated:
[tex]\[ t_1 \approx 0.45 \][/tex]
[tex]\[ t_2 \approx 0.19 \][/tex]
[tex]\[ t_3 \approx 0.14 \][/tex]
So, the ratio of the times to fall equal heights [tex]\( h \)[/tex] is approximately:
[tex]\[ t_1 : t_2 : t_3 = 0.45 : 0.19 : 0.14 \][/tex]
Hence, the ratio of the times is:
[tex]\[ \boxed{0.45 : 0.19 : 0.14} \][/tex]
### Step 1: Understand the Problem
A particle is released from rest from the top of a tower of height [tex]\( 3h \)[/tex]. We need to find the ratio of the times taken to fall each height [tex]\( h \)[/tex].
### Step 2: Equations of Motion
We will use the basic equation of motion for freely falling objects:
[tex]\[ h = \frac{1}{2} g t^2 \][/tex]
where:
- [tex]\( h \)[/tex] is the height,
- [tex]\( g \)[/tex] is the acceleration due to gravity ([tex]\( \approx 9.8 \, \text{m/s}^2 \)[/tex]),
- [tex]\( t \)[/tex] is the time taken to fall that height.
### Step 3: Time to Fall Each Height [tex]\( h \)[/tex]
We need to find the times [tex]\( t_1 \)[/tex], [tex]\( t_2 \)[/tex], and [tex]\( t_3 \)[/tex] to fall each height [tex]\( h \)[/tex].
#### Time [tex]\( t_1 \)[/tex] to fall the first height [tex]\( h \)[/tex]:
From [tex]\( 3h \)[/tex] to [tex]\( 2h \)[/tex], the particle falls the first height [tex]\( h \)[/tex].
Using the motion equation:
[tex]\[ h = \frac{1}{2} g t_1^2 \][/tex]
Solving for [tex]\( t_1 \)[/tex]:
[tex]\[ t_1 = \sqrt{\frac{2h}{g}} \][/tex]
#### Time [tex]\( t_2 \)[/tex] to fall the second height [tex]\( h \)[/tex]:
From [tex]\( 2h \)[/tex] to [tex]\( h \)[/tex], the particle falls the second height [tex]\( h \)[/tex]. This time is the difference between the time to fall [tex]\( 2h \)[/tex] and the time to fall [tex]\( h \)[/tex].
[tex]\[ 2h = \frac{1}{2} g t_{2, \text{total}}^2 \][/tex]
Solving for [tex]\( t_{2, \text{total}} \)[/tex]:
[tex]\[ t_{2, \text{total}} = \sqrt{\frac{4h}{g}} \][/tex]
The time [tex]\( t_2 \)[/tex] to fall the second height [tex]\( h \)[/tex] is:
[tex]\[ t_2 = t_{2, \text{total}} - t_1 \][/tex]
[tex]\[ t_2 = \sqrt{\frac{4h}{g}} - \sqrt{\frac{2h}{g}} \][/tex]
#### Time [tex]\( t_3 \)[/tex] to fall the third height [tex]\( h \)[/tex]:
From [tex]\( h \)[/tex] to [tex]\( 0 \)[/tex], the particle falls the third height [tex]\( h \)[/tex]. This time is the difference between the time to fall [tex]\( 3h \)[/tex] and the time to fall [tex]\( 2h \)[/tex].
[tex]\[ 3h = \frac{1}{2} g t_{3, \text{total}}^2 \][/tex]
Solving for [tex]\( t_{3, \text{total}} \)[/tex]:
[tex]\[ t_{3, \text{total}} = \sqrt{\frac{6h}{g}} \][/tex]
The time [tex]\( t_3 \)[/tex] to fall the third height [tex]\( h \)[/tex] is:
[tex]\[ t_3 = t_{3, \text{total}} - t_{2, \text{total}} \][/tex]
[tex]\[ t_3 = \sqrt{\frac{6h}{g}} - \sqrt{\frac{4h}{g}} \][/tex]
### Step 4: Find the Ratio [tex]\( t_1 : t_2 : t_3 \)[/tex]
Substituting the values we calculated:
[tex]\[ t_1 \approx 0.45 \][/tex]
[tex]\[ t_2 \approx 0.19 \][/tex]
[tex]\[ t_3 \approx 0.14 \][/tex]
So, the ratio of the times to fall equal heights [tex]\( h \)[/tex] is approximately:
[tex]\[ t_1 : t_2 : t_3 = 0.45 : 0.19 : 0.14 \][/tex]
Hence, the ratio of the times is:
[tex]\[ \boxed{0.45 : 0.19 : 0.14} \][/tex]
