Answer:
20.96 m/s
Explanation:
Using the equations of motion
y = uᵧt + gt²/2
Since the puck slides off horizontally,
uᵧ = vertical component of the initial velocity of the puck = 0 m/s
y = vertical height of the platform = 2 m
g = 9.8 m/s²
t = time of flight of the puck = ?
2 = (0)(t) + 9.8 t²/2
4.9t² = 2
t = 0.639 s
For the horizontal component of the motion
x = uₓt + gt²/2
x = horizontal distance covered by the puck
uₓ = horizontal component of the initial velocity = 20 m/s
g = 0 m/s² as there's no acceleration component in the x-direction
t = 0.639 s
x = (20 × 0.639) + (0 × 0.639²/2) = 12.78 m
For the final velocity, we'll calculate the horizontal and vertical components
vₓ² = uₓ² + 2gx
g = 0 m/s²
vₓ = uₓ = 20 m/s
Vertical component
vᵧ² = uᵧ² + 2gy
vᵧ² = 0 + 2×9.8×2
vᵧ = 6.26 m/s
vₓ = 20 m/s, vᵧ = 6.26 m/s
Magnitude of the velocity = √(20² + 6.26²) = 20.96 m/s
