At what angle with the horizontal should a hunter aim his gun so he will hit a target is 60 meters horizontally away from him, and 20.0 mm above him, if the muzzle velocity of the bullet is 400 m/s? This one's giving me a hard time.

I know x=ucos(theta)t
y= usin(theta)t-(gt^2)/2

but dang doing the algebra to solve for theta is a pain. Anybody got an idea how to make this easier?

Question

At what angle with the horizontal should a hunter aim his gun so he will hit a target is 60 meters horizontally away from him, and 20.0 mm above him, if the muzzle velocity of the bullet is 400 m/s? This one's giving me a hard time. 

I know x=ucos(theta)t 
y= usin(theta)t-(gt^2)/2 

but dang doing the algebra to solve for theta is a pain. Anybody got an idea how to make this easier?

anonymous · Answer

range or horizontal distance=60 = u^2sin2@/g

vertical displacement =20mm= (usin@)^2/2g

first before u do anything convert mm to METer

sin2@=2sin@*cos@
this wilol help u a lot
if u have doubt solving feel free to ask

anonymous · Answer

oops, sorry y should be 2.00 m not mm. Well I know the trig identities but here:$$y=xtan \theta-(gx^2)/2u^2\cos^2\theta$$ at some point I did get the sin2(theta) identity but the whole thing's far from being solved.

anonymous · Answer

equate both equation and solve for cos@

anonymous · Answer

Isn't $$x=(u^2\sin2\theta)/g$$ can only be used when the elevation of the initial and final positions in y the same? In this case though, that isn't the situation.

anonymous · Answer

no matter wat the H or vertical displacement gives tghe MAXIMUM HEIGHT reached by the projectile