Doing this problem?
If you're playing volleyball and one of the teammates serves the ball by hitting it from a height of 8 ft in arc that is parabolic, imagine that the ball reaches its max height of 12 ft, 4 ft to the right of where it was originally served. But after traveling 4 more feet after that, it gets stuck on the net.
1) What is the equation representing the trajectory of the volleyball if x and y represent horizontal and vertical distance from the start point?
2) How high is the net?
- Wayne DeguManLv 71 month ago
The physical method would be to set up equations involving gravity and time.
However, we can model on a quadratic function that relates x to y
so, y = -a(x - b)² + c
As this is in vertex form and we know (4, 12) gives the maximum value we have:
y = -a(x - 4)² + 12
Also, when x = 0, y = 8 so,
8 = -a(-4)² + 12
i.e. 8 = -16a + 12
a = 1/4
Hence, y = (-1/4)(x - 4)² + 12
When x = 4, the ball reaches the greatest height of 12 which is midway along it's horizontal flight.
Hence, if it travels another 4 feet, due to symmetry, it returns to the vertical level of projection.
so, height of net is 8 feet
Note, this is the same as saying, what is y when x = 8
=> y = (-1/4)(8 - 4)² + 12
i.e. y = (-1/4)(16) + 12 => 8
Going a bit further, it will hit the ground when:
0 = (-1/4)(x - 4)² + 12
Giving, x = 11.9 feet
A sketch is below.
- 1 month ago
x(t) = d/dx v(t) = d²/dx² a(t)
a(t) = -9.8
v(t) = -9.8t + C → C represents the initial velocity
x(t) = -4.9t² + Ct + D → D represents the initial height (8 ft)
Interpreting the max height:
"...reaches its max height of 12 ft..."
v(t) = 0 = -9.8t + C
t = C / 9.8
x(C/9.8) = 12 = -4.9(C/9.8)² + C(C/9.8) + 8
... Solve for C
x(t) = -4.9t² + Ct + 8 → Insert C to get your function
To find the height of the net, we know it travels 4 feet to the right after that. Since parabolas are symmetric, and the max height occurs 4 feet right of where it was served, just plug in:
x(2C/9.8) = -4.9(2C/9.8)² + C(2C/9.8) + 8
And you solved for C previously