Jose bautista hits a baseball that travels for 142m before it lands. The flight of the ball can be modelled by a quadratic function in which x is the Horizontal distance the ball has travelled away from jose, and h(x) is the height Vertical distance of the ball at that distance.
Assume that the ball was between 0.6m and 1.5m above the ground when it was hit.
1. determine an equation that models the path of the ball, given this additional info:
-the ball was 1.2 m off the ground when it was hit.
-the ball reached a max of 17m in height at approximately 70m away from Jose.
Explain the method you are using for this equation.
So I know that the vertex is (70, 17).
But I'm lost on how to do this equation.
Your vertex is indeed (70,17)
Other points would be (142,0) and (0,1.2)
Since the vertex must lie midway between the two intercepts and the distance between 70 and 142 is 72, the other intercept must be (-2,0)
h(x) = a(x-70)^2 + 17 must be the equation
use (142,0) to find a
0 = a(72)^2 + 17
a = -17/5184 = appr -.00327932
h(x) = -.00327932(x-70)^2 + 17
check: does (0,1.2) lie on it?
LS = 1.2
RS = -.00327932(-70)^2 + 17 = appr .93
Were were expecting 1.2 , which is between .9 and 1.5 as originally stated.
But then it also said: " the ball was 1.2 m off the ground when it was hit. " , which appears to contradict the results of my equation.
If we take (0, 1.2) as fact , let's use that to find a
1.2 = a(-70)^2 + 17
a = -15.8/4900 = appr -.0032245
in that case, if x = 142, we should get zero
h(0) = -.0032245(72)^2 + 17 = .28 , close but mathematically too great an error.
It appears to me that some of the data is slightly off.posted by Reiny
College of education Kangere
Mathematics & interscieceposted by sulaiman abdullahi