Aball is shot straight up from the surface of the
earth with an initial speed of 19.6 m/s. neglect
any effects due to air resistance.
29. what is the magnitude of the ball's displacement
from the starting point after 1.00 has elapsed?
30. what maximum height will the ball reach?
31. how much time elapses between the throwing
of the ball and its return to the original launch
point?

A ) The displacement:
d = v o t - (gt²) / 2 = 
= 19.6 m/s × 1 s - ( 9.8 m/s² x 1 s² ) / 2 = 
= 19.6 m - 4.9 m = 14.7 m
b ) v = v o - g t
0 = 19.6 m/s - 9.8 t    ( when the ball is at the highest point )
9.8 t = 19.6
t = 19.6 : 9.8
t = 2 s
h = v o t - (gt²)/2 = 19.6 x 2 - ( 9.8 x 4 ) / 2 = 39.2 - 19.6
h = 19.6 m
c ) h = gt² / 2
19.6 = 9.8 t²/2
9.8 t² = 39.2
t² = 39.2 : 9.8
t² = 4
t 2 = 2 s   ( and we know that t 1 = 2 s )
t = t 1 + t 2 = 2 s + 2 s = 4 s 

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