maximum heigth ball reaches

Question

the path of a baseball after it is hit is modelled by the equation h=-4.98x²=12.51t+1.36, where h is the height, in meters, 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?

 

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

Answer 1

You didn't write the equation correctly: h=-4.98x²=12.51t+1.36 doesn't really make sense

 

I'm guessing you meant h=-4.98t²+12.51t+1.36 -- is that correct?

 

Also, I assume you're studying calculus, so we can use calculus techniques for this, right?

 

If that's the case then...

 

a) 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 -9.8t+12.51.  So solve this equation for t:

 

-9.96t+12.51 = 0

12.51 = 9.8t

1.256 = t

 

Substitute this value of t into the original equation to get the maximum height.

 

b) Substitute 2.5 into the equation to get the height at 2.5 seconds.  HOWEVER...  If the value at 2.5 seconds comes out negative, then you can't use that value as the answer, because presumably the ball would land on the ground and not go any lower.  So in that case, the answer would be 0.  (Math is good for solving phyiscal problems, but you always have to make sure to take the physical logic of the situation into account in interpreting the results.)

 

c) Solve the equation -4.98t²+12.51t+1.36 = 8 using the quadratic formula.

 

d) 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.

 

Answer 2

Hi Layne,

 

I take it that you meant + when you typed =

 

h(t)=-4.98t²+12.51t+1.36

 

h^'(t)=-4.98*2t+12.51 = -9.96t+12.51 h^'(0)=12.51 maximum height (a)

 

h(2.5)=-4.98(2.5)²+12.51(2.5)+1.36 (b)

 

 

h(t) = 8 = -4.98t²+12.51t+1.36   t is equal to how much? Use quadratic equation.