A projectile is thrown at angle θ with an inclined plane of inclination 45o .
Find θ if projectile strikes the inclined the plane horizontal
Taking x-axis along the incline and y-axis perpendicular to incline.[/B]
Vx=ucosθ - gsint(45)t
Vy=usinθ - gcos(45)t
These are the velocities after time t.
Vx=Velocity along the plane after time t.
Vy=Velocity perpendicular to plane after time t.
The Attempt at a Solution
As at the time of horizontal collision with the incline. The projectile will make an angle of 45o with the incline and hence the velocity of projectile V=Vcos45i + Vsin45j which means the x and y components of velocity are equal (as sin45=cos45).
And hence I applied the condition that Vx=Vy
that gives us:
ucosθ - gsint(45)t = usinθ - gcos(45)t
=> ucosθ-usinθ = gsin(45)t-gcos(45)t
=>u(cosθ-sinθ) = gt(sin45-cos45)
=>u(cosθ-sinθ)=0 (as sin45=cos45)
=>cosθ=sinθ => Tanθ=1
which gives θ=45o
But the answer is θ = tan-1(2) - 45o
I know how the right solution came but I am more bothered about what was wrong in my attempt that I got it incorrect.
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