Question

An arrow is shot vertically upward and then 1.81 s later passes the top of a tree 43.9 m high. How much longer will the arrow travel upward, and how high will it go?

Answer 1

using 2nd equation of motion

d = ut - 0.5 gt^2

43.9 = u* 1.81 - 4.9* 1.81^2

u = 33.123 m/s

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a)

time of travel in upward direction

t = u/g = 33.123 / 9.8 = 3.3799 s

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b)

max height

h = u^2 / (2 g)

h = 33.123^2 / 19.6

h = 55.977 m

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