kyle drops a ball from a height of 15 feet above the ground. The function h = -16 f2 +15 gives the height h of the ball (in feet) after t seconds. Graph the function to determine at about what time the ball hits the ground
so, did you graph it?
i need helppp why is there no graph sorry if i sound rude :(
To determine at what time the ball hits the ground, we need to find when the height (h) is equal to zero.
The given function is: h = -16t^2 + 15
To graph this function, we can plot points for different values of t. Let's choose a few values for t and compute the corresponding heights:
For t = 0, h = -16(0)^2 + 15 = 15
For t = 1, h = -16(1)^2 + 15 = -1
For t = 2, h = -16(2)^2 + 15 = -47
For t = 3, h = -16(3)^2 + 15 = -97
Plotting these points on a graph, we have:
(0, 15)
(1, -1)
(2, -47)
(3, -97)
The graph will be a downward-opening parabola.
Now, let's plot the points on a coordinate plane and draw a smooth curve passing through them:
```
| *
| *
| /
| /
-------|------------------
t |
```
From the graph, we can see that the ball hits the ground between t = 1 and t = 2 seconds. To find a more accurate answer, we can use the concept of finding the x-intercept of the function. In this case, the x-intercept represents the time when the height is zero.
Setting h = 0 in the equation -16t^2 + 15 = 0:
-16t^2 + 15 = 0
16t^2 = 15
t^2 = 15/16
t = sqrt(15/16)
t ≈ 0.96875
Therefore, at about 0.96875 seconds, the ball hits the ground.
To determine at about what time the ball hits the ground, we need to find the value of t when the height h equals zero.
Given that the function for the height of the ball is h = -16t^2 + 15, we can set h = 0 and solve for t:
0 = -16t^2 + 15
To solve this quadratic equation, we can factor or use the quadratic formula.
Factoring method:
0 = -16t^2 + 15
16t^2 = 15
t^2 = 15/16
t = ±√(15/16)
Quadratic formula method:
Using the quadratic formula, t = (-b ± √(b^2 - 4ac)) / (2a), where a = -16, b = 0, and c = 15.
t = (0 ± √(0^2 - 4(-16)(15))) / (2(-16))
t = (0 ± √(0 + 960)) / (-32)
t = (0 ± √(960)) / (-32)
t = ± √(960) / (-32)
t = ± √(960) / (-32)
t ≈ ±3.09
Since time cannot be negative, we discard the negative value of t.
Therefore, the ball hits the ground at about t ≈ 3.09 seconds.