Answer to Question #202112 in Physics for xyjcai

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°:

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