Final answer:
The ball launched horizontally from 1.2 m high at 4 m/s will hit the ground in approximately 0.49 seconds. If launched at a 30° angle with the same speed from the same height, it will land approximately 1.70 meters away from the launcher.
Step-by-step explanation:
To determine the time it takes for the ball to hit the ground when launched horizontally from a height of 1.2 m with a speed of 4 m/s, we only need to consider the vertical motion, as the horizontal motion does not influence the time it takes for the projectile to reach the ground. The equation for the vertical motion is d = 0.5 * g * t^2, where d is the displacement (1.2 m downwards), g is the acceleration due to gravity (9.8 m/s^2), and t is the time in seconds. Solving for t, we find that t = sqrt(2 * d / g). Substituting the values, we get t = sqrt(2 * 1.2 m / 9.8 m/s^2), which calculates to approximately 0.49 seconds.
For the angled launch at 30° with the same speed of 4.00 m/s, we need to calculate both the horizontal and vertical components of the initial velocity. The horizontal component is Vx = V * cos(θ) and the vertical component is Vy = V * sin(θ), where V is the initial speed and θ is the launch angle. Once we have Vy, we can use the same vertical motion equation to find the time, and then use this time to calculate the range the projectile covers horizontally with range = Vx * time. The calculated time is the same as the previous part, 0.49 seconds. The horizontal range is then range = Vx * t = (4 m/s * cos(30°)) * 0.49 s, which results in a distance of approximately 1.70 meters from the launcher.