A) To find the maximum height reached by the ball, we need to determine the vertex of the quadratic function h(t) = -5t^2 + 10t + 40. The vertex of a quadratic function is given by the formula:
t = -b / (2a)
In this case, a = -5 and b = 10. Let's substitute these values into the formula:
t = -10 / (2 * (-5)) t = -10 / (-10) t = 1
So the ball reaches its maximum height at t = 1 second.
To find the maximum height, substitute t = 1 back into the equation:
h(1) = -5(1)^2 + 10(1) + 40 h(1) = -5 + 10 + 40 h(1) = 45
Therefore, the maximum height reached by the ball is 45 meters.
B) To determine when the ball hits the water, we need to find the time (t) when the height (h) is equal to zero. Set h(t) = 0 and solve for t:
-5t^2 + 10t + 40 = 0
We can solve this quadratic equation by factoring or by using the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -5, b = 10, and c = 40. Substituting these values into the formula:
t = (-10 ± √(10^2 - 4 * (-5) * 40)) / (2 * (-5)) t = (-10 ± √(100 + 800)) / (-10) t = (-10 ± √900) / (-10) t = (-10 ± 30) / (-10)
Now we have two possibilities:
- When t = (-10 + 30) / (-10) = 20 / (-10) = -2
- When t = (-10 - 30) / (-10) = -40 / (-10) = 4
Since time cannot be negative in this context, we discard the value t = -2.
Therefore, the ball hits the water at t = 4 seconds.