Question

A particle is projected with a velocity of 30m/s at an angle 60 to the horizontal.Find the range on the plane inclined at 30 to the horizontal when projected from a point on the plane and up the plane

Answer 1

Answer:

60 m

Explanation:

A particle is projected with a velocity of 30m/s at an angle 60 to the horizontal.Find the range on the plane inclined at 30 to the horizontal when projected from a point on the plane and up the plane

Horizontal Velocity = 30Cos60 = 15 m/s

Vertical Velocity = 30Sin60 = 15√3 m/s

Time to reach Peak

= 15√3/g = 15√3/10 = 2.6 sec

Horizontal Distance covered = 15 * 2.6 = 39 m

Vertical Distance covered = (15√3)²/2g = 33.75m

Let say after this t sec is taken

to hit inclined plane

Vertical Distance = 5t² (down wards)

Horizontal Distance = 15t

Height Above Ground  = 33.75 - 5t²

Horizontal Distance = 39 + 15t

=> Tan 30 = (33.75 - 5t²)/(39 + 15t)

=> 1/√3 = (33.75 - 5t²)/(39 + 15t)

=> 39 + 15t = 58.45 - 5√3t²

=> 5√3t² + 15t  - 19.45 = 0

=> t = 0.86 sec

Vertical Height Above Ground = 33.75 - 5t² = 30 m

Horizontal Distance = 51.9 m

Distance on inclined Planed = √(30)² +(51.9)² = 60 m