Question 6: For any specific velocity of projection, the range of projection cannot exceed from the value equal to four times of the corresponding height. Discuss.
Similarly, the horizontal range of the projectile is
Now from trigonometry, sin2θ = 2sinθcosθ, put in the above equation,
Multiply the RHS by
Rearrange and simplify,
Put value of H from equation (A)
Now for maximum range, Rm, θ =45°, therefore,
Answer
For any specific velocity of projection, maximum height of the projectile is,
Similarly, the horizontal range of the projectile is
Now from trigonometry, sin2θ = 2sinθcosθ, put in the above equation,
Multiply the RHS by
Rearrange and simplify,
Put value of H from equation (A)
Now for maximum range, Rm, θ =45°, therefore,
R tan45º = 4H ⇒ R = 4H
(Since tan 45° = 1)
Now as 45° gives the maximum range of a projectile for a specific velocity, therefore, 4H is the corresponding maximum height. This proves the statement.