SOLUTION: 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

Algebra ->  Graphs -> SOLUTION: 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      Log On


   

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.
b. Determine the height of the ball, to the nearest hundredth of a metre, at 2.5 seconds.
c. Determine when the ball is at a height of 8m, to the nearest hundredth of a second. (2 marks)
d. State the y-intercept of the graph to the nearest hundredth. Describe what the y-intercept represents in the context of this question.
e.i. Find the coordinates of both x-intercepts of the graph to the nearest hundredth.
ii. State what the positive x-intercept represents in the context of this problem.
iii. Explain why the negative x-intercept is not valid in the context of this problem.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

7. The path of a baseball after it is hit is modelled by the equation
h=-4.98t%5E2%2B12.51t%2B1.36, 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%2B12.51+=+0.........solve this equation for t
12.51+=+4.98t
12.51%2F+4.98=+t
t=2.51
Substitute this value of t into the original equation to get the maximum height.
h=-4.98%282.512%29%5E2%2B12.51%282.512%29%2B1.36
h=1.36

b. Determine the height of the ball, to the nearest hundredth of a metre, at 2.5 seconds.
h=-4.98%282.5%29%5E2%2B12.51%282.5%29%2B1.36
h=1.51m
c. Determine when the ball is at a height of 8m, to the nearest hundredth of a second. (2 marks)
8=-4.98t%5E2%2B12.51t%2B1.36
0=-4.98t%5E2%2B12.51t%2B1.36-8
0=-4.98t%5E2%2B12.51t-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%5E2%2B12.51t%2B1.36.........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.