Explanation:
To find an equation for the height of the ball, we can use the information provided: the initial height of the ball, the time it takes to reach the maximum height, and the maximum height itself.
Let's assume the equation for the height of the ball is h(t) = at^2 + bt + c, where h(t) represents the height at time t, and a, b, and c are constants to be determined.
Given the information:
When t = 0, the ball's height is 1.63 meters:
h(0) = a(0)^2 + b(0) + c = c = 1.63
When t = 0.6 seconds, the ball reaches the maximum height of 3.33 meters:
h(0.6) = a(0.6)^2 + b(0.6) + c = 3.33
To determine the remaining constants, we need one more piece of information. Since we don't have the exact height at that point, let's assume the ball is back at the initial height after 1.2 seconds (twice the time it took to reach the maximum height).
When t = 1.2 seconds, the height of the ball is back to 1.63 meters:
h(1.2) = a(1.2)^2 + b(1.2) + c = 1.63
Now we have a system of three equations:
c = 1.63 (equation 1)
0.36a + 0.6b + c = 3.33 (equation 2)
1.44a + 1.2b + c = 1.63 (equation 3)
Substituting equation 1 into equations 2 and 3, we get:
0.36a + 0.6b + 1.63 = 3.33 (equation 4)
1.44a + 1.2b + 1.63 = 1.63 (equation 5)
Simplifying equation 5, we find:
1.44a + 1.2b = 0
Now, we have two equations with two variables:
0.36a + 0.6b = 1.7 (equation 4)
1.44a + 1.2b = 0 (equation 5)
Solving this system of equations, we find:
a = -1.5
b = 2.5
Now we can substitute the values of a, b, and c back into the equation for the height of the ball:
h(t) = -1.5t^2 + 2.5t + 1.63
Therefore, the equation for the height, h, of the ball in meters after t seconds is h(t) = -1.5t^2 + 2.5t + 1.63.