Answer:
The correct numbers are;
1) 0
2) 25
3) 1.25
4) 2.5
When the ball is hit at 0 seconds, it has a height of 0 feet. The ball's height increases until it reaches its maximum height of 25 feet after 1.25 seconds. The ball's height then decreases until it reaches the ground 2.5 seconds after it was hit
Explanation:
The given equation for the height of the ball, h(t), in feet, is given as follows
h(t) = 40·t - 16·t²
Therefore;
1) When the ball is hit at 0 seconds, it has a height of h(0) = 40×0 - 16×0² = 0 feet
2) The time to reach maximum height by the ball is found by taking the derivative (slope) of the function for the height of the ball and noting that at maximum height, the derivative (slope) = 0
Therefore;
h'(t) = 40 - 2 × 16·t = 0
40 - 32·t = 0
t = 40/32 = 5/4 seconds = 1.25 seconds
3) The maximum height is then found from the value of the function at t = 5/4 seconds as follows;
h(5/4) = 40×5/4 - 16×(5/4)² = 50 - 25 = 25 feet
4) The time in which the ball reaches the ground where h(t) = 0 is given from the formula for the height of the gulf ball as follows;
h(t) = 0 = 40·t - 16·t²
40·t - 16·t² = 0
(40 - 16·t)·t = 0
Therefore;
t = 0 seconds or t = 40/16 = 2.5 seconds
(40 - 16·t) = 0/t = 0
t = 40/16 = 2.5 seconds
Therefore, the filling the blanks with the correct numbers gives;
When the ball is hit at 0 seconds, it has a height of 0 feet. The ball's height increases until it reaches its maximum height of 25 feet after 1.25 seconds. The ball's height then decreases until it reaches the ground 2.5 seconds after it was hit