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)

##19.4"m"##

(b)

##134.4"m"##

(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=19.4"m"##

(b)

The expression for range ##d## on level ground is:

##d=(v^2sin(2theta))/(g)##

##:.d=(39^2xxsin(60))/(9.8)##

##d=134.4"m"##