A helicopter drops a box of supplies after a disaster. The function represents the height y (in feet) of the box t seconds after it is dropped. After how many seconds does the box hit the ground?
y=100-16t^2
Your equation suggests that the box was dropped from a height of 100 ft
solve :
0 = 100-16t^2
16t^2 = 100
take square root of both sides
4t = 10
take over.
Well, let's see. The equation for the height of the box is y = 100 - 16t^2. Now, if the box hits the ground, it means that its height y is equal to 0. So, we can set up the equation 0 = 100 - 16t^2 and solve for t.
Let me calculate that for you...
*takes out a calculator and presses some imaginary buttons*
Ah, yes! The answer is quite simple. After how many seconds does the box hit the ground? Drumroll, please...
The box hits the ground after 2.5 seconds. So, if you want to catch it, you better be quick!
To determine when the box hits the ground, we need to find the value of t when the height y is equal to zero. In other words, we need to solve the equation 100 - 16t^2 = 0 for t.
First, let's rewrite the equation in a more familiar form:
16t^2 = 100
Next, divide both sides of the equation by 16:
t^2 = 100/16
Simplify the right side:
t^2 = 6.25
Now, take the square root of both sides, considering both positive and negative roots:
t = ±√6.25
In this case, negative time does not make sense, so we only consider the positive root.
Calculating the square root of 6.25 gives us:
t ≈ 2.5
Therefore, the box hits the ground approximately 2.5 seconds after it is dropped.
To find out when the box hits the ground, we need to determine the value of t when the height y is equal to zero. In this case, the box hits the ground when y = 0.
The given function is:
y = 100 - 16t^2
To solve for t when y = 0, we set the equation equal to zero:
0 = 100 - 16t^2
To simplify this equation, we need to isolate the variable t. Let's start by subtracting 100 from both sides:
-100 = -16t^2
Next, divide both sides by -16 to solve for t^2:
(t^2) = 100/16
Simplify the right side:
t^2 = 6.25
Now, take the square root of both sides to solve for t:
t = ±√(6.25)
Since time cannot be negative in this context, we take the positive square root:
t = √(6.25)
Evaluating the square root of 6.25, we get:
t ≈ 2.5
Therefore, the box hits the ground approximately 2.5 seconds after it is dropped.