You are target shooting using a toy gun that fires a small ball at a speed of 15 m/s. when the gun is fired at an angle of 31 degrees above horizontal, the ball hits the bull's-eye of a target at the same height as the gun. then the target distance is halved.

at what angle must you aim the gun to hit the bull's-eye in its new position? (mathematically there are two solutions to this problem; the physically reasonable answer is the smaller of the two.)

\theta 2=

13.1° from the horizontal

Explanation: For projectile

Horizontal distance R of a projectile is:

R= U×Usin2x/g

Where U is the speed of projectile, x is angle of projectile and g= acceleration due to gravity=9.8m/s

R= 15×15sin(2×31)/9.8= 225sin(62)/9.8=20.27m

Therefore if R is halved.

R/2 = 20.27/2=10.14m

Hence the angle for this case is.

Making sin(2x) the subject of formula

Sin2x= Rg/U×U

R is now 10.14 in this case.

Sin2x= 10.14×9.8/15×15=0.4415

2x=arcsin0.4415=26.2

x= 26.2/2

x= 13.1°

Good luck...

the answer is a. at the equator

i think they are all correct

explanation:

they seem good