a) s=ut+1/2at^2 where u is initial vertical velocity which equals zero because it is thrown horizontally
40=1/2(9.81)t^2
t^2=40/4.905
t= 2.85568...
t=2.86
b) R=vt
R=30x2.86
R=85.67m
c) 40-25 = 15
therefore, distance is 15m
s=ut+1/2at^2
15=1/2(9.81)t^2
t^2=15/4.905
t=1.7487...
t=1.75s
d)s=ut+1/2at^2
s=1/2(9.81)(2)^2
s=19.62
distance traveled after 2 sec is 19.62 metres...
therefore, height above ground: 40-19.62 = 20.38 metres above ground