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Answered

a ball 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.

  1. What is the magnitude of the ball's displacement

from the starting point after 1.00 has elapsed?

  1. what maximum height will the ball reach?

  2. how much time elapses between the throwing

of the ball and its return to the original launch


point?

Verified Answer :

HomertheGenius

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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