physics problem!?

Question

A small ball rolls off the edge of a tabletop that is 1.20m high. It strikes the floor at a point 1.52m horizontally from the table edge. a)how long is the ball in the air? b) what is its speed at the instant it leaves the table?

Answer 1

h=1.52m
g=10 m/s^2
h=u^2/2g
1.52=u^2/20
u^2=1.52*20=34
taking square on both sides wehave
u=square root of 34
u=5.83m/s
t=u^2/2g
t=u/g
t=5.83/10=0.583 sec
therefore
a) the ball remained 0.583 sec in air
b) instant speed is 5.83 m/s
important note the point where it fell shouild be considered as height but you told that it reached the point horizontally
so the height must be 1.20m but not 1.52 m your question might be wrong
if you take height as 1.20m calculation might be like this
h=1.20m
g=10m/s^2
h=u^2/2g
1.20=u^2/20
u^2=1.20*20
u^2=24
taking square on both sides
u= square root of 24=4.898m/s
t=u/g=4.898/10=0.4898sec
therefore
a)the ball remained 0.4898sec in air
b)its initial speed is 4.898m/s

Answer 2

The fall of the ball will follow the motion equation of
1.2=.5*9.81*t^2
t=0.4946 s

The horizontal motion will follow
1.52=vi*t
since t=0.4946
vi=1.52/0.4946
vi=3.07 m/s

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