Question

A ball is kicked into the air and follows the path described by h(t)= – 4.9t2 + 6t + 0.6, where t is the time in seconds and h is the height in meters above the ground. Find the maximum height of the ball. What value would you have to change in the equation to increase the maximum height of the ball?

Answer 1

Answer:

The maximum height of the ball is 2.437 meters. Change in initial value will  increase the maximum height of the ball.

Step-by-step explanation:

The given equation is

$h(t)=-4.9t^2+6t+0.6$

Where t is the time in seconds and h is the height in meters above the ground.

It is second degree polynomial with negative leading coefficient therefore, it shows a downward parabola. A downward parabola givens the maximum value at vertex.

If the parabola is defined as

$f(x)=ax^2+bx+c$

Then the vertex of the parabola is

$(-\frac{b}{2a},f( -\frac{b}{2a}))$

Since the value of a and b are -4.9 and 6 respectively.

$-\frac{b}{2a} =-\frac{6}{2(-4.9)} =0.612$

put this value in the given equation.

$h(t)=-4.9(0.612)^2+6(0.612)+0.6=2.437$

Therefore the vertex is (0.612,2.437) and the maximum height of the ball is 2.437 meters.

If we increase the value of a upto less than 0, b and c in the equation the height of ball increases. But when the the value of a is greater than 0, then the function will give minimum value and if x=0, then the function is a linear line.

For example we increase the initial height (c) upto 10. Then

$h(t)=-4.9t^2+6t+10$

The vertex of the function is (0.612,11.837). Now the maximum height of the ball is 11.837. Change in initial value will increase the maximum height of the ball.