Answer:
y=39.057 m
Explanation:
Using Kinematic relation
[tex]s=ut+ \frac{1}{2}at^2[/tex]
given u= 5m/s
a=g= -9.81 [tex]m/s^2[/tex]( directed downward)
[tex]s=5t- \frac{1}{2}(9.80)t^2[/tex]
Also, we know that
v=u+at
v=5-9.80t
at time t= 0.250 sec
[tex]s=5\times0.25- \frac{1}{2}(9.80)0.25^2[/tex]
s=0.94375 m
now position of sandbag
y= 40-0.94375
y=39.057 m