A particle is projected at an angle α with the horizontal from the foot of a plane, whose inclination to the horizontal is β. Find the velocity with which the particle strikes perpendicular to the inclined plane.
A
ucos(α−β)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
usin(α−β)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
usin(α2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
ucos(β2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Busin(α−β)
As the final velocity v is along Y-direction, we can say v=usin(α−β)−gcosβT where T is the time of flight. T=2usin(α−β)gcosβ ⇒v=usin(α−β)−gcosβ[2usin(α−β)gcosβ]∴v=−usin(α−β)
Thus, velocity with which particle strikes the plane is usin(α−β)