Final answer:
The maximum height attained by the ball is 36.4 m, and the horizontal distance from the point where the ball was thrown to the point where it strikes the roof is 141.2 m.
Step-by-step explanation:
To determine the maximum height attained by the ball, we need to analyze the motion of the ball. The initial vertical velocity of the ball is given by the product of the initial speed and the sine of the launch angle: 40 m/s * sin(60°) = 34.64 m/s. The time taken to reach the maximum height can be determined using the equation: t = (Vf - Vi) / a, where Vf = 0 m/s, Vi = 34.64 m/s, and a = -9.8 m/s^2 (acceleration due to gravity). Substituting in these values, we can find t = 3.53 s.
The maximum height attained by the ball can be calculated using the equation: h = Vi * t + (1/2) * a * t^2, where Vi = 34.64 m/s, t = 3.53 s, and a = -9.8 m/s^2. Substituting in these values, we find h = 34.64 m/s * 3.53 s + (1/2) * (-9.8 m/s^2) * (3.53 s)^2 = 36.4 m.
Therefore, the maximum height attained by the ball is 36.4 m (option D).
To determine the horizontal distance traveled by the ball, we can use the equation: d = V * t, where V = initial horizontal velocity of the ball and t = total time of flight. The initial horizontal velocity can be determined as the product of the initial speed and the cosine of the launch angle: 40 m/s * cos(60°) = 20 m/s. The total time of flight is twice the time taken to reach the maximum height: 2 * 3.53 s = 7.06 s. Substituting these values, we find d = 20 m/s * 7.06 s = 141.2 m.
Therefore, the horizontal distance from the point where the ball was thrown to the point where it strikes the roof is 141.2 m (option B).