Projectile motion with trig? (Gen Physics 1 question)

Question

An archer shoots an arrow at a target that is a horizontal distance d = 55 m away; the bull’s-eye of the target is at same height as the release height of the arrow.

At what angle, in degrees above the horizontal, must the arrow be released to hit the bull’s-eye if the arrow's initial speed is 39 m/s?

5686 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-22T16:21:31

A useful expression to use for the range is:

#sf(d=(v^2sin2theta)/g)#

#:.##sf(sin2theta=(dg)/(v^2))#

#sf(sin2theta=(55xx9.81)/39^2)#

#sf(sin2theta=0.3547)#

#sf(2theta=20.77^@)#

#sf(theta=10.4^@)#

Answer 2 · 2017-02-22T17:40:17

#15.65^@#

The parabolic path described by the arrow considering the coordinates origin at the archer position, is

#(x,y) = (v_0 cos theta t,v_0 sin thetat -1/2 g t^2)#

After #t_0# seconds the target is hit so

#v_0 cos theta t_0 = d-> t_0 = d/(v_0 cos theta)#

at this time #t_0# also

#v_0 sin theta t_0 -1/2 g t_0^2 = 0# or substituting

#v_0 sin theta (d/(v_0 cos theta))-1/2g (d/(v_0 cos theta))^2=0 #

Simplifying

#v_0^2sin theta cos theta-1/2gd^2=0# or

#2sintheta costheta = sin(2theta)=(g d)/v_0^2# and finally

#theta = 1/2 arcsin((gd)/v_0^2) = 1/2 arcsin(9.81(55/39^2)) = 0.273148#[rad] = #15.65^@#