Final answer:
The package strikes the ground 266 m away from the point directly below where it was released. The horizontal component of the velocity just before it hits is 38 m/s. The vertical component of the velocity just before it hits is 68.6 m/s.
Step-by-step explanation:
First, we need to find the time it takes for the package to hit the ground. We can use the equation y = 0.5 * g * t^2, where y is the height, g is the acceleration due to gravity, and t is the time. Plugging in the known values: y = -195 m, g = 9.8 m/s^2, we can solve for t. Rearranging the equation, we get t = sqrt(2 * y / g), so t = sqrt((2 * -195) / 9.8) = 7 s. Next, we can find the horizontal distance using the equation x = v * t, where x is the distance, v is the velocity, and t is the time. Plugging in the known values: v = 38 m/s, t = 7 s, we can solve for x. x = 38 m/s * 7 s = 266 m. Therefore, the package strikes the ground 266 m away from the point directly below where it was released.
To find the horizontal component of the velocity just before it hits, we can use the equation v = x / t, where v is the velocity, x is the distance, and t is the time. Plugging in the known values: x = 266 m, t = 7 s, we can solve for v. v = 266 m / 7 s = 38 m/s. Therefore, the horizontal component of the velocity just before it hits is 38 m/s.
To find the vertical component of the velocity just before it hits, we can use the equation v = g * t, where v is the velocity, g is the acceleration due to gravity, and t is the time. Plugging in the known values: g = 9.8 m/s^2, t = 7 s, we can solve for v. v = 9.8 m/s^2 * 7 s = 68.6 m/s. Therefore, the vertical component of the velocity just before it hits is 68.6 m/s.