Question
An arrow is shot straight upward into the air and it returns to the ground with a speed of 28m/s. What was the maximum height reached by the arrow?
Answers
Damon
v = Vi - 9.81 t
if it came down at 28 m/s, it started up at 28 m/s (symmetry)
v = 28 - 9.81 t
at the top, v = 0
0 = 28-9.81 t
so
t = 28/9.81 at the top, same time falling
h = (1/2)(9.81) t^2
h = (4.9)(28/9.81)^2
h = 4.9
if it came down at 28 m/s, it started up at 28 m/s (symmetry)
v = 28 - 9.81 t
at the top, v = 0
0 = 28-9.81 t
so
t = 28/9.81 at the top, same time falling
h = (1/2)(9.81) t^2
h = (4.9)(28/9.81)^2
h = 4.9