Menu Close

Problem 9: The maximum height gained by a projectile is 300 m. If it travels a range of 800 m, find the displacement of summit point from the point of projection.


The trajectory of the projectile can be depicted as in following diagram. 

ADC is the trajectory of the projectile with D as the highest point and AC as the range of the projectile. Now, AC = 800 m, DB = 300 m and AB = 400 m. Apply Pythagorean theorem to the right triangle,

(AD)2 = (AB)2 + (BD)2

Put the values,

(AD)2 = (400)2 + (300)2

(AD)2 = 160000 + 90000 = 250000

AD = √ 250000 = 500 m

This is the displacement of the summit (highest point) from the point of projection (A).

1 Comment

  1. Pingback:numerical-problems-force-and-motion-chapter-3-physics-11 – msa

Leave a Reply

Your email address will not be published. Required fields are marked *