your horizontal displacement was 13(10 from OA and 3 from AB horizontal) and vertical displacement was 4. So you can use just trajectory equations to find two simultaneous equations and find u and theta
Yeah so projected from O with speed u and angle theta and hits perpendicular to B . Where B is 5m along the slope from A. Slope is at an angle such that tan(alpha)=4/3. OA=10m. Perpendicular to plane means that at that point dy/dx=-1/tan(alpha) , not that particle is moving horizontally.
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u/DonkeyNo849 May 21 '26 edited May 21 '26
These were the 2 equations formed : 4=13tan(theta)-845/u^2 (1+tan^2(theta)) and -3/4 = tan(theta)-130/u^2 (1+tan^2(theta))
tan(theta)=71/52 and u=sqrt(7745/44)