A pole vaulter clears 6.00 m. With what velocity does the vaulter strike the mat in the landing area? (Disregard air resistance. g = 10 m/s2.) 2gh = v2

a
10.8 m/s
b
21.6 m/s
c
2.70 m/s
d
5.40 m/s

1 answer

To calculate the velocity with which the pole vaulter strikes the mat, we will use the conservation of energy principle, specifically considering the potential energy at the highest point and the kinetic energy just before hitting the ground.

Given:

  • Height (h) = 6.00 m
  • Acceleration due to gravity (g) = 10 m/s²

We can use the formula:

\[ v^2 = 2gh \]

Substituting the values:

\[ v^2 = 2 \cdot 10 , \text{m/s}^2 \cdot 6.00 , \text{m} \]

Calculating:

\[ v^2 = 2 \cdot 10 \cdot 6 = 120 , \text{m}^2/\text{s}^2 \]

Now, taking the square root to find \(v\):

\[ v = \sqrt{120} \approx 10.95 , \text{m/s} \]

Rounding it off, we find:

\[ v \approx 10.8 , \text{m/s} \]

Thus, the velocity with which the vaulter strikes the mat is approximately \( \mathbf{10.8 , m/s} \).

The correct answer is a) 10.8 m/s.