Question

Derive an expression for time of flight of projectile motion

Answer 1

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 ⁡ . The angle of reach is the angle the object must be launched at in order to achieve a specific distance: θ=12sin−1(gdv2) θ = 1 2 sin − 1 ⁡ ( gd v 2 ).

Answer 2

Answer:

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 ⁡ . The angle of reach is the angle the object must be launched at in order to achieve a specific distance: θ=12sin−1(gdv2) θ = 1 2 sin − 1 ⁡ ( gd v 2 ) .