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Question 11: For any specific velocity of projection, the range of projection cannot exceed from the value equal to four times of the corresponding height. Discuss.


We know that for any specific velocity v and angle θ of projection, the maximum height of projectile is

Similarly, the horizontal range of the projectile is

Since sin2θ = 2sinθcosθ. Put in the above equation for R and simplify,

Multiply and divide RHS by (sinθ)/2

The quantity in the brackets is height H of the projectile (equation A above), therefore,

Now cosθ/sinθ = cotθ = 1/tanθ. Put

Now for maximum range, Rm, θ =45°, therefore,

(This is because tan45° = 1)

Therefore, for any specific velocity, the range of a projectile can’t exceed than 4 times the maximum height of the projectile.

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  1. Pingback:Conceptual Questions Forces and Motion, Physics 11 – msa

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