Answer:
An object launched from 65° will have the same range as an object launched from 25°
Step-by-step explanation:
The formula for the horizontal range, R, is given as follows;
![R = (v * sin(2\cdot \theta))/(g)](https://img.qammunity.org/2021/formulas/physics/high-school/rviofoq98d50jk2srspdabljm1e5hsbobq.png)
Therefore, given that sinθ increases from 0 to 1 for θ values from 0 to 90° and decreases from 1 to 0 for θ values from 90° to 180°, and that the values of sin(θ) are positive in the first and second quadrant, we have that for the given angle, θ = 65°, the range is therefore;
![R = (v * sin(2* 65 ^(\circ)))/(g) = (v * sin(130 ^(\circ)))/(g)](https://img.qammunity.org/2021/formulas/physics/high-school/jrs6bhrvq0js14kc810iqksekhxjhtvmn4.png)
The location of 130° = Second quadrant
A rotation of 130° in the counterclockwise is equivalent to a rotation of 50° in the clockwise direction
Therefore, we have sin(130°) = sin(50°)
Therefore, the range for 130°/2 = 65° is the same as the range for 50°/2 = 25° and an object launched from 65° will have the same range as an object launched from 25°.