Question

a bullet is fired from the bottom of the inclined plane at an angle 37 degree with inclined plane the angle of incline is 30 with the horizontal .find the position of max height of the bullet from the inclined plane

Answer 1

Answer:

0.21V²/g

Explanation:

A bullet is fired from the bottom of the inclined plane at an angle 37 degree with inclined plane the angle of incline is 30 with the horizontal .find the position of max height of the bullet from the inclined plane

Horizontal Speed = Vcos(30 +37) = 0.39V

Vertical Speed = Vsin67 = 0.92V

Time to reach Max height = 0.92V/g

Horizontal Distance = 0.39V * 0.92V/g  = 0.36V²/g

Vertical Distance = (0.92V)²/2g = 0.42V²/g

Tan30 = P/B

=> 1/√3 = Incline plane vertical height / (0.36V²/g)

=>  Incline plane vertical height =  0.36V²/g√3

the position of max height of the bullet from the inclined plane

= 0.42V²/g - 0.36V²/g√3

= 0.42V²/g - 0.21V²/g√3

= 0.21V²/g