Asked by sebenano
A ball is thrown upwards and reaches the ground once again after 4 sec: (a)calculate the height that the ball reaches (b)the total distance covered by the ball (c)the velocity with which the ball hits the ground
Answers
Answered by
bobpursley
h(t)=h(0)+vi*t-1/2 g t^2
0=0+vi*4-4.9(4^2)
so you need the initial velocity, vi, or vi=4.9*4 m/s.
height: vi*2
distance: 2*vi*2
vf=-vi
0=0+vi*4-4.9(4^2)
so you need the initial velocity, vi, or vi=4.9*4 m/s.
height: vi*2
distance: 2*vi*2
vf=-vi
Answered by
Refilwe
Height=364.56m
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