Question

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=

Answer 1

Answer: 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...

Answer 2

The angle required to hit the bull's-eye in the new position is 13°

Determination of the range

From the question given above, the following data were obtained:

• Initial velocity (u) = 15 m/s
• Angle (θ) = 31°
• Acceleration due to gravity (g) = 9.8 m/s²

• Range (R) =?

R = u²Sine2θ / g

R = 15² × Sine(2×31) / 9.8

R = 225 × Sine 62 / 9.8

R = 20.27 m

How to determine the angle required to hit the bull's-eye in the new position

• Initial velocity (u) = 15 m/s
• Acceleration due to gravity (g) = 9.8 m/s²
• Range (R) = 20.27 / 2 = 10.135 m

• Angle (θ) =?

R = u²Sin2θ / g

10.135 = 15² × Sine2θ / 9.8

10.135 = 225 × Sine2θ / 9.8

Cross multiply

225 × Sine2θ = 10.135 × 9.8

225 × Sine2θ = 99.323

Divide both side by 225

Sine2θ = 99.323 / 225

Sine2θ = 0.4414

Take the inverse of Sine

2θ = Sine¯¹ 0.4414

2θ = 26

Divide both side by 2

θ = 26 / 2

θ = 13°

Learn more about projectile motion:

https://brainly.com/question/20814243