Question

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

Answer 1

The distance covered by an object starting from rest
and accelerating uniformly is . . .

                     Distance  =  (1/2 of acceleration) x (time)²

                                       = (1/2 · 9.8 m/s²)  x  (3.5 s)²

                                       =      (4.9 m/s²)  x  (3.5 s)²

                                        =      (4.9 m/s²)  x  (12.25 s²)

                                       =        (4.9 x 12.25)   meters

                                       =            60.025  meters