Final answer:
The horizontal distance a ball travels when launched at a 70° angle and reaching a height of 8.0 m before it has completely horizontal velocity cannot be determined without additional information about the initial velocity.
Step-by-step explanation:
To determine the horizontal distance from the launcher to the side of the building where a ball is launched at a 70° angle and lands with a horizontal velocity, it's important to understand projectile motion and the independence of horizontal and vertical motions. Since we know the ball reaches a height of 8.0 m and it has a horizontal velocity at that point, we can use the kinematic equation for vertical motion to find the time it took for the ball to reach that height (assuming initial vertical velocity is 0 at that point).
Using the equation √y = Viy • t + 0.5 • g • t2, where √y is the change in vertical position, Viy is the initial vertical velocity, g is the acceleration due to gravity (9.81 m/s2), and t is the time. Solving for t when the ball reaches the peak (Viy = 0) gives us √y = 0.5 • g • t2, hence t = √(2√y/g). Using the given height (8.0 m), t is approximately 1.28 s. Since we are looking for the horizontal distance and the horizontal velocity (Vix) is constant, we multiply this time by Vix, which is not given directly.
However, since the ball's velocity at the roof's edge is completely horizontal, the only force acting on the ball after it leaves the launcher but before reaching the roof's edge is gravity. This means we can ignore air resistance and calculate the horizontal distance using the time the ball takes to reach the peak, which is the same amount of time it would take to fall from that height. As a result, the horizontal distance traveled would be Vix • t. But as Vix is not specified, we cannot calculate that without additional information. Therefore, with the information provided, it's not possible to accurately pick one of the options as the final answer to the horizontal distance.