A ball thrown vertically upward is caught by
the thrower after 2.12 s.
Find the initial velocity of the ball. The
acceleration of gravity is 9.8 m/s
Answer in units of m/s.Find the maximum height it reaches
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ball thrown from yo= o, caught at the same spot, then y= 0
t= 2.12 s.
y = y0 + v0t – ? gt2
let submit soon pliz
44.o452/2.12 = 20.77
maximum height will be
time will be 2.12/2= 1.06
y = y0 + ? (v + v0)t = 0 + ?(0 + 20.77m/s)(1.06 s)
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