To solve this problem, we can use the principles of projectile motion. Let's break down the given information and solve for the time and horizontal distance.
a) Time required for the ball to fall to the ground:
We'll consider the vertical motion of the ball. The initial vertical velocity (Vy) is 0 since the ball is initially at rest in the vertical direction. The acceleration due to gravity (g) is approximately 9.8 m/s². The vertical displacement (Δy) is -0.60 meters (negative because the ball is falling downward).
Using the kinematic equation:
Δy = Vy * t + (1/2) * g * t²
Plugging in the values:
-0.60 = 0 * t + (1/2) * 9.8 * t²
-0.60 = 4.9 * t²
Solving for t, we have:
t² = -0.60 / 4.9
t² ≈ -0.1224
Since time cannot be negative, we discard the negative square root. Therefore,
t ≈ √(-0.1224) (ignoring the negative square root)
The time required for the ball to fall to the ground is approximately:
t ≈ 0.35 seconds
b) Horizontal distance between the table's edge and the ball's landing location:
We'll consider the horizontal motion of the ball. The initial horizontal velocity (Vx) is given as 8.64 km/hr. We need to convert it to m/s.
1 km/hr = 1000 m/3600 s = 5/18 m/s
8.64 km/hr = 8.64 * (5/18) m/s ≈ 2.4 m/s
Since there is no horizontal acceleration (assuming no air resistance), the horizontal velocity remains constant. We can use the equation:
Distance (d) = Vx * t
Plugging in the values:
d = 2.4 * 0.35
d ≈ 0.84 meters
The horizontal distance between the table's edge and the ball's landing location is approximately 0.84 meters.