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Refer to the diagram below.
Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².
The pole vaulter drops with an initial vertical speed u = 0.
At impact with the pad, velocity v satisfies:
v² = 2 × (9.8 m/s²) × (4.2 m) = 82.32 (m/s)²
v = 9.037 m/s
As the pad compresses by 0.5 m to bring the vaulter to rest,
let the average acceleration (deceleration) be a m/s². Then:
0 = (9.037 m/s)² + 2 × a × 0.5 m
Solving for a gives:
a = - 82.32 / (2 × 0.5) = -82 m/s²
Thus, the deceleration magnitude is 82 m/s².
Answer:
1.5 × 10³⁶ light-years
Explanation:
A particular square area in interstellar space measures roughly 2.4 × 10⁷² (light-years)². To find the area of a square, the following formula is utilized:
A = l²
where,
A represents the area of the square
l denotes the length of one side of the square
Thus, l = √A = √2.4 × 10⁷² (light-years)² = 1.5 × 10³⁶ light-years
The acceleration of an object will always align with the direction of the resultant force acting upon it. Thus, we can find horizontal acceleration by examining the horizontal force applied, applying Newton's second law in its mathematical form: Force = mass * acceleration. Therefore, acceleration = force / mass. By inputting the numbers, we have a = 100 / 0.15, which results in a = 666.7 m/s². Consequently, the acceleration experienced by the hockey puck is 670 m/s².