Answer:
7.35 m/s
Explanation:
Using y - y' = ut - 1/2gt², we find the time it takes the ball to fall from the 1.2 m table top and hit the floor.
y' = initial position of ball = 1.2 m, y = final position of ball = 0 m, u = initial vertical velocity of ball = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for ball to hit the ground.
So, substituting the values of the variables into the equation, we have
y - y' = ut - 1/2gt²
0 - 1.2 m = (0 m/s)t - 1/2(9.8 m/s²)t²
- 1.2 m = 0 - (4.9 m/s²)t²
- 1.2 m = - (4.9 m/s²)t²
t² = - 1.2 m/- (4.9 m/s²)
t² = 0.245 s²
t = √(0.245 s²)
t = 0.49 s
Since d = vt where d = horizontal distance ball moves = 3.6 m, v = horizontal velocity of ball = unknown and t = time it takes ball to land = 0.49 s.
So, d = vt
v = d/t
= 3.6 m/0.49 s
= 7.35 m/s
Since the initial velocity of the ball is 7.35 m/s since the initial vertical velocity is 0 m/s.
It is shown thus V = √(u² + v²)
= √(0² + v²)
= √(0 + v²)
= √v²
= v
= 7.35 m/s
