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A particle is projected from point A perpendicular to inclined plane with a velocity of 50 m/s. Particle strikes a vertical plane perpendicularly at point B. Find the time taken by particle in going from point A to point B.
as you said its is projected perprendicularly then angle becomes 90 therfore t=2usin90/g=2*50/10=10m/s
t=(50sin57)/10= 4.2s ~ 4 sat pt b velocity becomes perpendicular i.e it is at maximum hieght so t=(usinθ)/gfrom geometry θ =57 ,g=10, u=50.
Here, u = 50 m/secAngle(Ѳ) = 90og = 9.8 m/sec2But practically angle is 57 degreeThen T = 2usin Ѳ/g = 2x50x sin57/9.8 = 4.2 sec = approx 4 sec
You missed angle of inclination i.e. 37° (given)Practically angle of projection = 90-37 =53°As it hit B perpendicularly hence B is Hmax .Time Time taken for Hmax = usintheta/g = 50* 4/5 * 1/10 = 4 sec
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