Astone is dropped from the upper observation deck of a tower, 950 m above the ground. (assume g = 9.8 m/s2.) (a) find the distance (in meters) of the stone above ground level at time t.

d = 950 m - 4.9t^2 m The distance an object moves under constant acceleration is d = 0.5at^2 where d = distance a = acceleration t = time. Since we're falling and since we're starting at 950 m above ground, the formula becomes: d = 950 m - 0.5at^2 Substituting known values, and simplifying gives us d = 950 m - 0.5*9.8 m/s^2 * t^2 d = 950 m - 4.9 m/s^2 * t^2 Since time is in seconds, we can cancel out the seconds in the units, getting d = 950 m - 4.9t^2 m