Question

a ball thrown vertically upward with a speed of 10.6 m/sec from a topof a tower return to the earth in 6 sec . find the height of the tower

Answer 1

speed = v = 10.6 m/s

time = t = 6 s

distance = height = d = ?

ball travel double distance from earth to top of the tower and tower to earth so total distance is 2d

so,

v = 2d/t

vt = 2d

vt/2 = d

10.6×6/2 = d

10.6×3 = d

31.8 m = d


so the height of the tower is 31.8 metre

Answer 2

u=10.6 m s^-1
v=0 m s^-1
t=6 sec
from 1st equation of motion
v=u+at
0=10.6+a×6
-10.6=6a
a=-10.6/6
a=-1.76
from 3rd equation of motion
v^2-u^2=2as
(0)^2-(10.6)^2=2×-1.76×s
0-112.36=-3.52s
112.36=3.52s
s=112.36/3.52
s=31.92m
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