Question

The figure below shows the path of a projectile motion.

a)Obtain the expressions for maximum height and time of flight.

b)What is the angle of projection for maximum horizontal range?​

Answer 1

Explanation:

a) The maximum height of an object, given the initial launch angle and initial velocity is found with:h=v2isin2θi2g h = v i 2 sin 2 ⁡ θ i 2 g . The time of flight of an object, given the initial launch angle and initial velocity is found with: T=2visinθg T = 2 v i sin ⁡ .

b) Hence, at θ=45o , Range is maximum.