Prove that for angle of projection, which exceed or fall short of 45° by equal amount the ranges are equal.



Given:

Angle = θ =  45°


To Prove:

Exceed or fall short of 45° by equal amount the ranges are equal



Solution:

Exceed of 45°

Let θ = 45° + ф, The Range R₁

R₁ = vi²sin(2θ₁)g

R₁ = vi²sin(2(45°+ф))g

R₁ = vi²sin(90°+2ф)g

R₁ = vi²sin(90°+2ф)g

But sin (90° + 2ф) = cos 2ф hence

R₁ = vi²cos2фg ------------(1)

Fall short of 45°

Let θ = 45° - ф, The Range R₂

R₂ = vi²sin(2θ₂)g

R₂ = vi²sin(2(45°-ф))g

R₂ = vi²sin(90°-2ф)g

R₂ = vi²sin(90°-2ф)g

But sin (90° - 2ф) = cos 2ф hence

R₂ = vi²cos2фg ------------(2)

By comparing equation (1) an (2)

R₁ = R₂

Hence the required proof


************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK)     

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.