A particle is projected from the bottom of an inclined plane of angle 30∘ with a velocity of 20m/s at an angle of 45∘ with the horizontal. Find the horizontal distance travelled by the particle before it hits the plane.
A
x=80√3[√3−1]m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
x=40√3[√3−1]m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
x=403[√3−1]m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
x=803[√3−1]m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Bx=40√3[√3−1]m Horizontal distance travelled by the particle is x=Rcosθ, where R is the range of a projectile on an inclined plane.
We know R=u2gcos2θ[sin(2a−θ)−sinθ]⇒R=20210×cos230∘[sin(2×45∘−30∘)−sin(30∘)]⇒R=40010×34[√32−12]⇒R=803[√3−1]