Answer:
V = 20.96 m/s
The magnitude of the velocity of the puck just before it touches the ground is 20.96 m/s
Explanation:
For the vertical component of velocity;
Using the equation of motion;
vᵥ^2 = u ^2 + 2as
Where;
vᵥ = final vertical speed
u = initial vertical speed = 0
(The puck only have horizontal speed.)
a = g =acceleration due to gravity = 9.8m/s^2
s = vertical distance covered = 2.0m
initial horizontal speed = 20 m/s
The equation becomes;
vᵥ^2 = u ^2 + 2gs
Substituting the given values, we have;
vᵥ^2 = 0 + 2(9.8×2)
vᵥ^2 = 39.2
vᵥ = √39.2
vᵥ = 6.26 m/s
For the horizontal component of velocity;
Given that effect of the air resistance is negligible, then the final horizontal speed equals the initial horizontal speed since there is no acceleration.
vₕ = uₕ = 20 m/s
The resultant magnitude of velocity can be derived by;
V = √(vₕ² + vᵥ²)
V = √(20² + 6.26²)
V = 20.96 m/s
The magnitude of the velocity of the puck just before it touches the ground is 20.96 m/s
