A ball is projected perpendicularly from an inclined plane of inclination θ from horizontal, with velocity ′u′ as shown in figure. The time after which the projectile is making angle 45∘ with the inclined plane is
A
ugsinθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
ugcosθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
ug(sinθ+cosθ)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
ug(sinθ−cosθ)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Cug(sinθ+cosθ)
Taking X&Y axis along the inclined plane and perpendicular to the inclined plane respectively as shown in figure. ax=+gsinθ,ay=−gcosθ,uy=+u,ux=0 ⇒At anytime ′t′ the component of velocity vector, vx=ux+axt=0+gsinθ×t ∴vx=gsinθt...(i) vy=uy+ayt=+u−gcosθ×t vy=u−gcosθt...(ii) ⇒At anytime t the angle (α) made by velocity vector from the X−axis is given by: ∴tanα=vyvx tanα=u−gcosθtgsinθt tan45∘=u−gcosθtgsinθt ⇒gsinθt+gcosθt=u