If a ball dropped from a tower reaches the ground after 3.5 seconds, what is the height of the tower? Given: g = –9.8 meters/second2

60.025m.

Explanation:

S= ut + at^2/2 (2nd equation of motion)

S= 0 + (9.8)(3.5)^2 /2 (free fall case, initial velocity = 0)

S = 4.9 x 12.25

S= 60.025 m.

Disclaimer: did math in my head, so you better double check with a calculator.

answer; how far out of place which the object finished relative to the starting location;