\(h(5) = -4.9(25) + 49(5) = -122.5 + 245 = 122.5\) meters - High Altitude Science

Understanding the Context

This expression models a physical scenario involving vertical motion under constant gravitational acceleration. Specifically, it approximates the height ( h(n) ) (in meters) of an object at time ( n ) seconds, assuming:

• Initial upward velocity related to coefficient ( 49 )
  - Acceleration due to gravity approximated as ( -4.9 , \ ext{m/s}^2 ) (using ( g pprox 9.8 , \ ext{m/s}^2 ) divided by ( 2 ))
  - Time ( n ) given as 5 seconds

Step-by-Step Breakdown: ( h(5) = -4.9(25) + 49(5) )

1. Evaluate ( -4.9 \ imes 25 ):
   [
   -4.9 \ imes 25 = -122.5
   ]

Key Insights

2. Evaluate ( 49 \ imes 5 ):
   [
   49 \ imes 5 = 245
   ]

3. Add the results:
   [
   -122.5 + 245 = 122.5
   ]

Thus, at ( n = 5 ) seconds, the height ( h(5) = 122.5 ) meters.

Why Is This Significant?

This calculation illustrates how functions model real-world motion. The quadratic form ( h(n) = -4.9n^2 + 49n ) reflects the effect of gravity pulling objects downward while an initial upward velocity contributes height. At ( n = 5 ), despite gravity’s pull reducing height, the object’s momentum or initial push results in a still significant upward displacement of 122.5 meters.

Final Thoughts

Practical Applications

Understanding such equations helps in physics, engineering, and sports science, especially when predicting trajectories:

• Projectile motion of balls, rockets, or drones
  - Engineering simulations for optimal release angles and velocities
  - Field calculations in sports analytics, such as estimating ball height in basketball or volleyball

Conclusion

The calculation ( h(5) = -4.9(25) + 49(5) = 122.5 ) meters is not just a computation—it represents a meaningful physical quantity derived from a real-world model. By breaking down each term and recognizing the underlying motion, learners and professionals alike can apply these principles confidently in various scientific and technical fields.