thanks for the reply
ok... so the vertical component will be 60sinθ
v=u+at so 0= 60sinθ - 9.8t
so 1/2t= 60sinθ/9.8
so t = 60sinθ/4.9
the h velocity will be 60cosθ
therefore 3600sinθcosθ/4.9 = 50
so 734.7sinθcosθ = 50
sinθcosθ = 0.068
if this is correct then what do i do to find θ?
An archer fires an arrow at a target placed 50 m away. The centre of the target is at the same vertical height as the bow when the arrow is fired.
The arrow leaves the bow with velocity u = 60 m/s
Calculate the angle θ at which the arrow is fired, giving your answer in degrees.v=u+at
s=ut+1/2...