Question 1178347


 7. The path of a baseball after it is hit is modelled by the equation

 {{{h=-4.98t^2+12.51t+1.36}}}, where {{{h}}} is the height, in metres, of the ball and{{{ t}}} is the time, in seconds, after the ball is hit.

a. Find the maximum height that the ball reaches, to the nearest hundredth of a metre.
You have to take the derivative of the function and set it equal to{{{ 0}}}.  That's where the ball would reach it's top point.  So the derivative is  
 
{{{-4.98t+12.51 = 0}}}.........solve this equation for {{{t}}}
{{{12.51 = 4.98t}}}
{{{12.51/ 4.98= t}}}
 {{{t=2.51}}}
Substitute this value of t into the original equation to get the maximum height.

{{{h=-4.98(2.512)^2+12.51(2.512)+1.36}}}
{{{h=1.36}}}


b. Determine the height of the ball, to the nearest hundredth of a metre, at {{{2.5}}} seconds.

{{{h=-4.98(2.5)^2+12.51(2.5)+1.36}}}
{{{h=1.51}}}m

c. Determine when the ball is at a height of {{{8m}}}, to the nearest hundredth of a second. (2 marks)

{{{8=-4.98t^2+12.51t+1.36}}}
{{{0=-4.98t^2+12.51t+1.36-8}}}
{{{0=-4.98t^2+12.51t-6.64}}}.........using the quadratic formula we get
{{{t=0.76}}} seconds or{{{ t=1.75}}} seconds


d. State the y-intercept of the graph to the nearest hundredth. Describe what the y-intercept represents in the context of this question.

The y-intercept occurs when {{{t=0}}}.  So substitute {{{0}}} into the equation.  You get {{{1.36}}}.  What that means is that the ball STARTS at a height of {{{1.36}}} meters, which would be the height of the bat when it hits the ball.

y-intercept is at ({{{0}}}, {{{1.36}}})

e.i. Find the coordinates of both x-intercepts of the graph to the nearest hundredth.

set {{{h=0}}}
{{{0=-4.98t^2+12.51t+1.36}}}.........using the quadratic formula we get
{{{t=-0.10}}} or  
{{{t=2.62}}}
x-intercept is at ({{{-0.10}}}, {{{0}}}) and  ({{{2.62}}}, {{{0}}}) 

ii. State what the positive x-intercept represents in the context of this problem.

Time the ball hits the ground  {{{2.62}}} seconds after being hit.

iii. Explain why the negative x-intercept is not valid in the context of this problem.

The ball is starting from a height of {{{1.36}}} m at {{{t=0}}}.  Where it was before being hit is non-sequitur.