Question

How will you compare the range (horizontal distance) travelled by the

ball at angles of 25º and 65º? How about at angles of 40º and 50º?

Answer 1

We know that according to the equation

$$R=\frac{v^2\sin(2\theta)}{g}$$

the maximum range is at 45°. Compare the ranges at 25° and 65°:

$$R(25°)=\frac{v^2\sin(2·25°)}{g}=0.766\frac{v^2}{g},\\\space\\ R(65°)=\frac{v^2\sin(2·65°)}{g}=0.766\frac{v^2}{g},\\\space\\ R(25°)=R(65°).$$

For 40° and 50°:

$$R(40°)=\frac{v^2\sin(2·40°)}{g}=0.985\frac{v^2}{g},\\\space\\ R(50°)=\frac{v^2\sin(2·50°)}{g}=0.985\frac{v^2}{g},\\\space\\ R(40°)=R(50°).$$