Answer :
To determine the time it takes for a ball to reach its maximum height when thrown upwards and to find the maximum height it reaches, we can use the kinematic equations of motion. Let's go through the problem step by step:
### Given:
- Initial velocity, [tex]\( u = 15 \, \text{m/s} \)[/tex]
- Initial height, [tex]\( h = 100 \, \text{m} \)[/tex]
- Acceleration due to gravity, [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex]
### Step 1: Time to Reach Maximum Height
At maximum height, the final velocity ([tex]\( v \)[/tex]) is 0.
We use the first kinematic equation:
[tex]\[ v = u - g \cdot t \][/tex]
Setting [tex]\( v = 0 \)[/tex], we get:
[tex]\[ 0 = u - g \cdot t \][/tex]
[tex]\[ t = \frac{u}{g} \][/tex]
Substituting the given values:
[tex]\[ t = \frac{15}{9.81} = 1.53 \, \text{seconds} \][/tex]
So, it takes 1.53 seconds to reach the maximum height.
### Step 2: Calculate Maximum Height
We use the second kinematic equation to find the maximum height:
[tex]\[ h_{\text{max}} = h + u \cdot t - \frac{1}{2} g \cdot t^2 \][/tex]
Substituting the known values:
[tex]\[ h_{\text{max}} = 100 + 15 \cdot 1.53 - \frac{1}{2} \cdot 9.81 \cdot (1.53)^2 \][/tex]
Breaking it down:
[tex]\[ h_{\text{max}} = 100 + 22.95 - 11.48 \][/tex]
[tex]\[ h_{\text{max}} = 111.47 \, \text{meters} \][/tex]
So, the maximum height reached is 111.47 meters.
Therefore, the time to reach maximum height and the maximum height are:
[tex]\[ \boxed{1.53 \, \text{seconds}; 111.47 \, \text{meters}} \][/tex]
Hence, the correct answer is:
[tex]\[ \text{B 1.53 seconds; 111.48 meters} \][/tex]
(Note: There is a minor discrepancy in decimal rounding between 111.47 meters and 111.48 meters. For alignment purposes with the provided options, we select 111.48 meters.)
### Given:
- Initial velocity, [tex]\( u = 15 \, \text{m/s} \)[/tex]
- Initial height, [tex]\( h = 100 \, \text{m} \)[/tex]
- Acceleration due to gravity, [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex]
### Step 1: Time to Reach Maximum Height
At maximum height, the final velocity ([tex]\( v \)[/tex]) is 0.
We use the first kinematic equation:
[tex]\[ v = u - g \cdot t \][/tex]
Setting [tex]\( v = 0 \)[/tex], we get:
[tex]\[ 0 = u - g \cdot t \][/tex]
[tex]\[ t = \frac{u}{g} \][/tex]
Substituting the given values:
[tex]\[ t = \frac{15}{9.81} = 1.53 \, \text{seconds} \][/tex]
So, it takes 1.53 seconds to reach the maximum height.
### Step 2: Calculate Maximum Height
We use the second kinematic equation to find the maximum height:
[tex]\[ h_{\text{max}} = h + u \cdot t - \frac{1}{2} g \cdot t^2 \][/tex]
Substituting the known values:
[tex]\[ h_{\text{max}} = 100 + 15 \cdot 1.53 - \frac{1}{2} \cdot 9.81 \cdot (1.53)^2 \][/tex]
Breaking it down:
[tex]\[ h_{\text{max}} = 100 + 22.95 - 11.48 \][/tex]
[tex]\[ h_{\text{max}} = 111.47 \, \text{meters} \][/tex]
So, the maximum height reached is 111.47 meters.
Therefore, the time to reach maximum height and the maximum height are:
[tex]\[ \boxed{1.53 \, \text{seconds}; 111.47 \, \text{meters}} \][/tex]
Hence, the correct answer is:
[tex]\[ \text{B 1.53 seconds; 111.48 meters} \][/tex]
(Note: There is a minor discrepancy in decimal rounding between 111.47 meters and 111.48 meters. For alignment purposes with the provided options, we select 111.48 meters.)
