To solve the given problem, we need to use the provided formula for the velocity of an object at impact with the ground:
[tex]\[ v = \sqrt{2gh} \][/tex]
Where:
- [tex]\( v \)[/tex] is the velocity at impact,
- [tex]\( g \)[/tex] is the acceleration due to gravity,
- [tex]\( h \)[/tex] is the initial height.
We need to rearrange the formula to solve for the initial height [tex]\( h \)[/tex]:
[tex]\[ v^2 = 2gh \][/tex]
[tex]\[ h = \frac{v^2}{2g} \][/tex]
### Part (a)
Given:
- Velocity at impact [tex]\( v = 38 \)[/tex] ft/sec,
- Acceleration due to gravity [tex]\( g = 32 \)[/tex] ft/sec².
We need to find the initial height [tex]\( h \)[/tex]:
[tex]\[ h = \frac{v^2}{2g} \][/tex]
Substitute the given values:
[tex]\[ h = \frac{38^2}{2 \times 32} \][/tex]
First, calculate [tex]\( 38^2 \)[/tex]:
[tex]\[ 38^2 = 1444 \][/tex]
Then, calculate the denominator [tex]\( 2 \times 32 \)[/tex]:
[tex]\[ 2 \times 32 = 64 \][/tex]
Now, divide the results:
[tex]\[ h = \frac{1444}{64} \][/tex]
This simplifies to:
[tex]\[ h \approx 22.56 \text{ feet (rounded to the nearest hundredth)} \][/tex]
### Part (b)
Given:
- Velocity at impact [tex]\( v = 16 \)[/tex] m/sec,
- Acceleration due to gravity [tex]\( g = 9.8 \)[/tex] m/sec².
We need to find the initial height [tex]\( h \)[/tex]:
[tex]\[ h = \frac{v^2}{2g} \][/tex]
Substitute the given values:
[tex]\[ h = \frac{16^2}{2 \times 9.8} \][/tex]
First, calculate [tex]\( 16^2 \)[/tex]:
[tex]\[ 16^2 = 256 \][/tex]
Then, calculate the denominator [tex]\( 2 \times 9.8 \)[/tex]:
[tex]\[ 2 \times 9.8 = 19.6 \][/tex]
Now, divide the results:
[tex]\[ h = \frac{256}{19.6} \][/tex]
This simplifies to:
[tex]\[ h \approx 13.1 \text{ meters (rounded to the nearest tenth)} \][/tex]
Therefore, the solutions are:
- Part (a): The initial height is [tex]\( 22.56 \)[/tex] feet.
- Part (b): The initial height is [tex]\( 13.1 \)[/tex] meters.
