An arrow is shot at a 30 degree angle with a velocity of 39m/s. How high will it go? What horizontal distance will it travel?

(a)
##19.4m##
(b)
##134.4m##
(a)
Considering the vertical component ot the motion:
##v^2=u^2+2as##
This becomes:
##v^2=u^2-2gh##
##:.0=(39sin30)^2-2xx9.8xxh##
##:.0=380-19.6xxh##
##h=280/19.6=19.4m##
(b)
The expression for range ##d## on level ground is:
##d=(v^2sin(2theta))/(g)##
##:.d=(39^2xxsin(60))/(9.8)##
##d=134.4m##