Question

A hockey puck slides off the edge of a platform with an initial velocity of 20 m/s horizontally. The height of the platform above the ground is 2.0 m. What is the magnitude of the velocity of the puck just before it touches the ground

Answer 1

V = 20.96 m/s

The magnitude of the velocity of the puck just before it touches the ground is 20.96 m/s

Step-by-step 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

Answer 2

20.96 m/s

Step-by-step 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