You can
put this solution on YOUR website!a cricket ball moves according to the rule
d= 1 + 3/5x -1/50 x^2
where d is the height it rises after travelling x metrs horizontally
.
d= -1/50 x^2 + 3/5x + 1
.
ii) what is the maximum height
Since the coefficient associated with the x^2 term is negative, we know that it is a parabola that opens downwards. Finding the vertex gives you the maximum height.
x-axis of symmetry is at -b/2a
-b/2a = -(3/5)/(-2/50)
-b/2a = (3/5)/(1/25)
-b/2a = (3/5)(25/1)
-b/2a = (3/1)(5/1)
-b/2a = 15
To find the maximum height, plug it into:
d= -1/50 x^2 + 3/5x + 1
d= -1/50 (15)^2 + 3/5(15) + 1
d= -1/50 (225) + 45/5 + 1
d= -4.5 + 9 + 1
d = 5.5 meters
.
iii)what is its original height
Original height is when x=0
d= -1/50 x^2 + 3/5x + 1
d= -1/50 0^2 + 3/5(0) + 1
d= 1 meter
.
iv) if it was caught 2m above the ground, how far was it hit
Here, set d=0 and solve for x:
d= -1/50 x^2 + 3/5x + 1
2= -1/50 x^2 + 3/5x + 1.18
0= -1/50 x^2 + 3/5x - 2
Use the quadratic equation to solve for x doing so yields:
x = {3.82, 26.18}
We can throw out the 3.86 meters (too fast after being hit) leaving us with
x = 26 meters
.
(see details below)
| Solved by pluggable solver: SOLVE quadratic equation with variable |
Quadratic equation  (in our case  ) has the following solutons:
For these solutions to exist, the discriminant  should not be a negative number.
First, we need to compute the discriminant :  .
Discriminant d=0.2 is greater than zero. That means that there are two solutions:  .
Quadratic expression  can be factored:
Again, the answer is: 3.81966011250105, 26.1803398874989.
Here's your graph:
 |
