An object is dropped from a height of 256 feet. The height [tex]\( h \)[/tex] of the object after [tex]\( t \)[/tex] seconds is given by the equation [tex]\( h = 256 - 16t^2 \)[/tex].

1. When will the height be 192 feet?
[tex]\(\square\)[/tex] seconds

2. When will the object reach the ground?
[tex]\(\square\)[/tex] seconds



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.