Answer :
Alright, let's solve this problem step by step.
### 1. When will the height be 192 feet?
We start with the given height equation:
[tex]\[ h = 256 - 16t^2 \][/tex]
We need to find the time [tex]\( t \)[/tex] when the height [tex]\( h \)[/tex] is 192 feet. Let's set [tex]\( h \)[/tex] to 192 and solve for [tex]\( t \)[/tex]:
[tex]\[ 192 = 256 - 16t^2 \][/tex]
First, rearrange the equation to bring all terms involving [tex]\( t \)[/tex] to one side:
[tex]\[ 256 - 192 = 16t^2 \][/tex]
[tex]\[ 64 = 16t^2 \][/tex]
Next, divide both sides by 16 to isolate [tex]\( t^2 \)[/tex]:
[tex]\[ t^2 = \frac{64}{16} \][/tex]
[tex]\[ t^2 = 4 \][/tex]
To find [tex]\( t \)[/tex], we take the square root of both sides:
[tex]\[ t = \sqrt{4} \][/tex]
[tex]\[ t = 2 \text{ or } t = -2 \][/tex]
However, since time cannot be negative, we discard the negative solution:
[tex]\[ t = 2 \][/tex]
So, the object will be at a height of 192 feet after:
[tex]\[ \boxed{2 \text{ seconds}} \][/tex]
### 2. When will the object reach the ground?
To find out when the object reaches the ground, set the height [tex]\( h \)[/tex] to 0:
[tex]\[ 0 = 256 - 16t^2 \][/tex]
Rearrange the equation to solve for [tex]\( t \)[/tex]:
[tex]\[ 256 = 16t^2 \][/tex]
Next, divide both sides by 16:
[tex]\[ t^2 = \frac{256}{16} \][/tex]
[tex]\[ t^2 = 16 \][/tex]
Take the square root of both sides to find [tex]\( t \)[/tex]:
[tex]\[ t = \sqrt{16} \][/tex]
[tex]\[ t = 4 \text{ or } t = -4 \][/tex]
Again, since time cannot be negative, we discard the negative solution:
[tex]\[ t = 4 \][/tex]
So, the object will reach the ground after:
[tex]\[ \boxed{4 \text{ seconds}} \][/tex]
In summary:
- The object will be at 192 feet after 2 seconds.
- The object will reach the ground after 4 seconds.
### 1. When will the height be 192 feet?
We start with the given height equation:
[tex]\[ h = 256 - 16t^2 \][/tex]
We need to find the time [tex]\( t \)[/tex] when the height [tex]\( h \)[/tex] is 192 feet. Let's set [tex]\( h \)[/tex] to 192 and solve for [tex]\( t \)[/tex]:
[tex]\[ 192 = 256 - 16t^2 \][/tex]
First, rearrange the equation to bring all terms involving [tex]\( t \)[/tex] to one side:
[tex]\[ 256 - 192 = 16t^2 \][/tex]
[tex]\[ 64 = 16t^2 \][/tex]
Next, divide both sides by 16 to isolate [tex]\( t^2 \)[/tex]:
[tex]\[ t^2 = \frac{64}{16} \][/tex]
[tex]\[ t^2 = 4 \][/tex]
To find [tex]\( t \)[/tex], we take the square root of both sides:
[tex]\[ t = \sqrt{4} \][/tex]
[tex]\[ t = 2 \text{ or } t = -2 \][/tex]
However, since time cannot be negative, we discard the negative solution:
[tex]\[ t = 2 \][/tex]
So, the object will be at a height of 192 feet after:
[tex]\[ \boxed{2 \text{ seconds}} \][/tex]
### 2. When will the object reach the ground?
To find out when the object reaches the ground, set the height [tex]\( h \)[/tex] to 0:
[tex]\[ 0 = 256 - 16t^2 \][/tex]
Rearrange the equation to solve for [tex]\( t \)[/tex]:
[tex]\[ 256 = 16t^2 \][/tex]
Next, divide both sides by 16:
[tex]\[ t^2 = \frac{256}{16} \][/tex]
[tex]\[ t^2 = 16 \][/tex]
Take the square root of both sides to find [tex]\( t \)[/tex]:
[tex]\[ t = \sqrt{16} \][/tex]
[tex]\[ t = 4 \text{ or } t = -4 \][/tex]
Again, since time cannot be negative, we discard the negative solution:
[tex]\[ t = 4 \][/tex]
So, the object will reach the ground after:
[tex]\[ \boxed{4 \text{ seconds}} \][/tex]
In summary:
- The object will be at 192 feet after 2 seconds.
- The object will reach the ground after 4 seconds.
