What is the angle at which the archer fired the arrow?

[matty796]

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 at^2
v^2=u^2+2as

 

[LowlyPion]

Welcome to PF.

You have two components of velocity don't you?

Maybe first calculate the time it will be in the air in terms of the vertical component? Up and down to end at the target, so twice the up sounds good? That will give you the vertical component of velocity in 1 equation related to time.

Now with the time expressed as the vertical component, won't the horizontal velocity times that time end at the target 50 m away at the instant you want?

 

[matty796]

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?

 

[LowlyPion]

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?

Without doing the math it looks OK.

Recognize now that 2*sinθcosθ = sin(2*θ) ... [double angle identity]
(See ... http://www.sosmath.com/trig/douangl/douangl.html )

hence just take the arc sine ... sin^-1(2*.068) = 2*θ

Just divide by 2 for your initial angle θ.

 

[matty796]

funny thing is that i knew that rule but didn't think to apply it!
but all the same that questions been bugging me for days, thanks a lot.