Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A particle is projected from a point on the surface of a smooth inclined plane simultaneously another particle q is released on

IMG_20190123_090703.jpg A particle is projected from a point on the surface of a smooth inclined plane simultaneously another particle q is released on the smooth inclined plane from the same position p and q polite after t is equal to 4 seconds the speed of projection of P is

Answers (1)

best_answer

@Arvind gupta 

For Q

U=0,  a=gsin\Theta

So s_q=\frac{1}{2}gsin\Theta\ \ t^2=8gsin\Theta

Now for P

if the angle of projection of P with the incline surface is α 
=> ucosα = speed along the plane, a = gsinθ° = acceleration along the plane 

Similarly usinα = speed perpendicular  to the plane, a = gcosθ° = acceleration perpendicular to the plane 

time of flight (For P) = 2usinα/gcosθ° = 4 s 

usinα = 2gcosθ°

distance travelled down incline = s = ut + ½at² 
Sp= 4ucosα + 8gsinθ° 

So Sp=Sq

= 4ucosα + 8gsinθ° = 8gsinθ° 
=> cosα = 0 or α = 90°

So we get u = 2gcosθ°=

u=2gcos\Theta =2g\frac{1}{2}=g=10m/s^2

g stn6 gsin6 t2 = 8gsin6 2 2gcos6 = g = 10m/s2 2
Posted by

avinash.dongre

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

Jharkhand CM inaugurates free coaching institute for tribal medical, engineering students
anam-khan

JEE Main 2026: IIT Roorkee’s ECE emerges as top choice after CSE; closing rank rises
anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2

Latest Question