Answer:
- v(t) = 38 -9.8t
- h(t) = -4.9t² +38t +1.8
Explanation:
You want functions for the velocity v(t) and height h(t) of a ball thrown upward at 38 m/s from a height of 1.8 m. The acceleration is -9.8 m/s².
Relations
This sort of problem is usually discussed in a physics course. The relation between acceleration, velocity, and position is ...
velocity is the rate of change of position
acceleration is the rate of change of velocity
Then the inverse relations are ...
h(t) = ∫v(t)dt +h(0)
v(t) = ∫a(t)dt +v(0)
Application
When position is measured in one dimension (up = positive), these have the standard formulas:
v(t) = g·t + v(0)
h(t) = 1/2g·t² +v(0)·t +h(0)
Filling in the given values (g = -9.8, v(0) = 38, h(0) = 1.8), we have ...
v(t) = -9.8t +38 . . . . . . . . . . . . meters per second
h(t) = -4.9t² +38t +1.8 . . . . . . meters