Asked by Anonymous
A ball is thrown upward with an initial velocity of 4ft/sec from a height of ten feet. what is the maximum height that the ball will reach.
h(t)= -16+vt+s
Answers
Answered by
Damon
v = Vi - g t
at top v = 0
0 = 4 -32 t
t = 1/8 second
(4 ft /sec is really slow. I bet you mean 40)
h = 10 + 4 (1/8) - 16 (1/64)
= 10 + .5 - .25
= 10.25 ft
at top v = 0
0 = 4 -32 t
t = 1/8 second
(4 ft /sec is really slow. I bet you mean 40)
h = 10 + 4 (1/8) - 16 (1/64)
= 10 + .5 - .25
= 10.25 ft
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