a ball rolls off the edge of a tabletop 1m above the floor and strikes the floor at a point 1.5m horizontally from the edge of the table. Find the (a) time of flight (b) initial velocity, (c) final velocity just before it strikes the floor. Plsssss help

boi this answer makes no sense

tell us answer in more creative way .............. this solution makes no sense

To find the time of flight, initial velocity, and final velocity of the ball, we can use the laws of motion and the principles of projectile motion. Here's how you can solve this problem step by step:

Step 1: Identify the known values:
- Initial vertical displacement (h) = 1 m
- Horizontal displacement (range) (R) = 1.5 m
- Acceleration due to gravity (g) = 9.8 m/s² (assumed to be near the surface of the Earth)

Step 2: Determine the time of flight (T):
To find the time of flight, we need to calculate the time it takes for the ball to reach the floor. In projectile motion, the vertical displacement is given by the equation:
h = (1/2)gt²

Plugging in the known values:
1 = (1/2)(9.8)t²

Solving for t:
t² = 2/9.8
t ≈ 0.45 seconds

Therefore, the time of flight is approximately 0.45 seconds.

Step 3: Calculate the initial velocity (Vi):
The initial velocity can be found using the horizontal displacement (range) and the time of flight. In projectile motion, the range (R) is given by the equation:
R = VixT

Rearranging the equation to solve for Vi:
Vi = R / T

Plugging in the known values:
Vi = 1.5 / 0.45
Vi ≈ 3.33 m/s

Therefore, the initial velocity is approximately 3.33 m/s.

Step 4: Determine the final velocity (Vf):
Since the ball strikes the floor, the final vertical velocity will be downwards. We can calculate it using the equation:
Vf = Vi + gt

Plugging in the known values:
Vf = 3.33 + (9.8)(0.45)
Vf ≈ 7.47 m/s

Therefore, the final velocity just before it strikes the floor is approximately 7.47 m/s.

To summarize:
(a) The time of flight is approximately 0.45 seconds.
(b) The initial velocity is approximately 3.33 m/s.
(c) The final velocity just before it strikes the floor is approximately 7.47 m/s.

time to fall?

h = (1/2) g t^2
so
1 = 4.9 t^2 solve for t

then
d = u t
1.5 = u t solve for u

also need v at floor
v = g t = 9.81 t

speed = sqrt(u^2 + v^2)
angle down from horizontal = tan^-1(v/u)