Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Live 1-1 coding classes to unleash the creator in your Child
A particle is dropped from height 100m and another particle is projected vertically upward with velocity 50m/s from the ground along the same line.find out the position where two particle will meet
I'm doing this ques with the concept of relative motionLet us consider dropped particle A and upward thrown particle B.(suppose u are a insect and sitting on B , now B will be at rest for u)
Now ,Vab = Va-Vb (velocity of A w.r.t B) (while sitting on A seeing B is Vab)Vab = 0-50 = -50m/s aab= -g-(-g) =0 ( acceleration of A w.r.t B)Distance=100m (as time and distance between 2 particle are independent of frame) Now , since a = 0 it is a simple non-accelerated motion. Applying ,Speed=distance/time 50=100/t=>time=2 sec Now, we can calculate distance travelled by A in time=2sec , that is at which height they will meet.) =>S=ut+at2/2 ( 2nd equation of motion)=>S=(0)(2)-(10)(2)2/2=>S=-20m ( negative sign signifies direction) So, the particles will meet 20m from top or 30m above ground. Here is ur requested answer, there is also a suggestion if u understood just do it without seeing above
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -