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Grade: 11


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.

5 years ago

Answers : (4)

Dhulipala Srikanth
45 Points
 as you said its is projected perprendicularly then angle becomes 90 therfore t=2usin90/g=2*50/10=10m/s
5 years ago
23 Points
t=(50sin57)/10= 4.2s ~ 4 s
at pt b velocity becomes perpendicular i.e it is at maximum hieght so t=(usinθ)/g
from geometry θ ​=57 ,g=10, u=50.
3 years ago
Megha Jain
24 Points
Here, u = 50 m/sec
Angle(Ѳ) = 90o
g = 9.8 m/sec2
But practically angle is 57 degree
Then T = 2usin Ѳ/g
             = 2x50x sin57/9.8
             = 4.2 sec
            = approx 4 sec
3 years ago
Hemant Jha
9 Points
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
2 years ago
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