Final answer:
To solve this problem of projectile motion, we can use equations of motion. The time of flight of the projectile can be determined using the horizontal distance traveled and the equation x = vt. The original velocity can be found using the equation v = d/t, and the velocity in the horizontal direction when it strikes the floor is the same as the original velocity. The velocity in the vertical direction at this time can be found using the equation vf = vi + at, and the magnitude of the velocity as it strikes the floor can be found using the equation v = sqrt(vx^2 + vy^2).
Step-by-step explanation:
In order to solve this problem, we can use the equations of motion for projectile motion. First, let's find the time of flight (a). Since the horizontal distance traveled is 2.050 m and the vertical distance fallen is 0.450 m, we can use the equation x = vt, where x is the horizontal distance, v is the horizontal velocity, and t is the time of flight. Plugging in the values, we get 2.050m = v*t. Next, let's find the original velocity (b). Since the horizontal velocity is constant throughout the motion, the original velocity is the same as the horizontal velocity. Using the equation v = d/t, where v is the horizontal velocity, d is the distance, and t is the time of flight, we can solve for v. Plugging in the values, we get v = 2.050m / t.
Finally, let's find the velocity in the horizontal direction when it strikes the floor (c). Since the horizontal velocity is constant throughout the motion, the velocity in the horizontal direction when it strikes the floor is the same as the original velocity. Therefore, the answer is the same as the answer to (b). Now, let's find the velocity in the vertical direction at this time (d). At the time when it strikes the floor, the vertical velocity is equal to the final velocity in the vertical direction.
The final velocity in the vertical direction can be found using the equation vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time of flight. Since the vertical distance fallen is 0.450 m and the acceleration due to gravity is 9.8 m/s^2, we can rearrange the equation to solve for vf. Plugging in the values, we get vf = sqrt(vi^2 + 2ad).
Finally, let's find the magnitude of its velocity as it strikes the floor (e). The magnitude of the velocity can be found using the equation v = sqrt(vx^2 + vy^2), where v is the magnitude of the velocity, vx is the horizontal velocity, and vy is the vertical velocity. Since we already found the values of vx and vy in (b) and (d), we can plug in the values to find the magnitude of the velocity.