JEE MainPhysicsModerateProjectile Motion

Two inclined planes OA and OB of inclinations to the horizontal
$\alpha\ and\ \beta$ , each equal to 30º are placed as shown in the
figure. A particle is projected at an angle of 90º with plane OA from
point A and its strikes the plane OB at point B normally. If the speed
of projection in m/s is x. Fill the value of 8 x. (given that OA=OB=20
cm and $g = 10m/s^{2})$

Show answer and solution

Since

\alpha = \beta

AO = BO = 20cm = d

cos120^{\circ} = \frac{d^{2} + d^{2} - (AB)^{2}}{2d^{2}}

AB = \sqrt{3}d

Now ABis the range of projectile from A to B

\sqrt{3}d = \frac{u^{2}sin2\theta}{g}

u = 2m/s

Source: Grass-G / P-16 Physics paper 25 Nov First.docx

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.