A bird drops a stick to the ground from a height of 68ft. The function h=-16t^2+68 gives the stick's approximate height h above the ground, in feet after t seconds. At about what time does the stick hit the ground
The stick will hit the ground when h = 0
- 16 t² + 68 =0
Multiply both sides by - 1
16 t² - 68 =0
Add 68 to both sides
16 t² = 68
t² = 68 / 16 = 4 ∙ 17 / 4 ∙ 4 = 17 / 4
t = ± √17 / √4 = ± √17 / 2
Time can't be be negative so:
t = √17 / 2
t = 2.0615528128 sec
Thank you so much:)
Oh, so the stick has been dropped by a bird, huh? I guess you could say the bird is really "branching" out with its stick-dropping skills! Anyway, let's solve this problem, shall we?
To find the time when the stick hits the ground, we need to set the height h equal to zero and solve for t in the equation h = -16t^2 + 68. So, let's plug in h = 0:
0 = -16t^2 + 68
To make our lives easier, let's divide both sides by 4:
0 = -4t^2 + 17
Now, let's subtract 17 from both sides:
-17 = -4t^2
Dividing by -4 (and remember, when you divide an inequality by a negative number, the inequality flips!), we get:
4.25 = t^2
Taking the square root of both sides, we find:
t ≈ ±2.06
Since time can't be negative in this context, we'll just take the positive value. So, approximately at t ≈ 2.06 seconds, the stick hits the ground. Just be careful not to be under it when it happens!
To find out at what time the stick hits the ground, we need to determine when the height (h) becomes zero.
The given function is h = -16t^2 + 68.
Setting h = 0, we get:
0 = -16t^2 + 68
Rearranging the equation:
16t^2 = 68
Dividing by 16:
t^2 = 68/16
Simplifying:
t^2 = 4.25
Taking the square root of both sides:
t = √4.25
Calculating the square root:
t ≈ 2.06
Therefore, at about 2.06 seconds, the stick hits the ground.
To find the time at which the stick hits the ground, we need to find the value of t when the height (h) is equal to zero.
Given the function h = -16t^2 + 68, we can set it equal to zero and solve for t:
-16t^2 + 68 = 0
Now, we can solve this quadratic equation for t. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -16, b = 0, and c = 68. Plugging these values into the formula, we have:
t = (0 ± √(0^2 - 4(-16)(68))) / (2(-16))
Simplifying further:
t = ± √(0 - (-4352)) / (-32)
t = ± √(4352) / (-32)
To find a positive value for t (since time cannot be negative), we take the positive square root:
t ≈ √(4352) / (-32)
Calculating the approximate value:
t ≈ 9.8777 / (-32)
t ≈ -0.3087
Since time cannot be negative, we discard the negative value and take the positive value as:
t ≈ 0.309 seconds
Therefore, the stick hits the ground at approximately 0.309 seconds.