To find the maximum height reached by the ball, we need to determine the vertex of the parabolic function h(t) = -5t^2 + 10t + 40. The vertex of a parabola in the form y = ax^2 + bx + c is given by the coordinates (h, k), where h = -b/2a and k = f(h).
In this case, a = -5, b = 10, and c = 40. Let's calculate the time when the ball reaches its maximum height:
h = -b/2a = -10 / (2 * -5) = -10 / -10 = 1
Now, substitute t = 1 into the equation to find the maximum height:
h(1) = -5(1)^2 + 10(1) + 40 = -5 + 10 + 40 = 45
Therefore, the maximum height reached by the ball is 45 meters.
To determine when the ball hits the water, we need to find the time at which h(t) becomes zero (since the water level is at h = 0). We can solve the equation -5t^2 + 10t + 40 = 0 using the quadratic formula.
The quadratic formula is given by:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
In our case, a = -5, b = 10, and c = 40. Substituting these values into the quadratic formula:
t = (-10 ± sqrt(10^2 - 4 * -5 * 40)) / (2 * -5) = (-10 ± sqrt(100 + 800)) / (-10) = (-10 ± sqrt(900)) / (-10) = (-10 ± 30) / (-10)
This gives two solutions:
t1 = (-10 + 30) / (-10) = 20 / (-10) = -2 t2 = (-10 - 30) / (-10) = -40 / (-10) = 4
Since time cannot be negative in this context, we discard t1 = -2. Therefore, the ball hits the water after 4 seconds.
To summarize: A) The maximum height reached by the ball is 45 meters. B) The ball hits the water after 4 seconds.