physics ball is thrown

Question

In a hypothetical universe, the magnitude of acceleration of an object subject to the gravitational force field of an Earth-like planet increases with the amount of time in seconds, t , spent in the field as

a = (9.8 m/s2) (1 - e^-0.1t)

A ball is thrown vertically upward at 40m/s. What is the final position after 1 second of travel? How fast is it moving at that moment? Assume that the origin is at the initial launch position.

Answer 1

Ball,s distance from the origin is y(t) = 40t - 4.9t² and velocity is v(t) = 40 - 9.8t

So, at t = 1s, y(1) = 40 - 4.9 = 35.1 m; v(1) = 40 - 9.8 = 30.2 m/s