Question

A particle P is thrown vertically upwards at speed 30 ms-1 from the ground. At the
same time, another particle Q, which is vertically above P, is released from rest at a
height of 120 m from the ground. Determine the time taken by P to meet Q

Answer 1

let p and q meet at distance x
then distance covered by q=x m
and by p = 120 -x m
then by formula s = ut +1/2 at^2
for p
s=120-x=1/2gt^2 (acceleration is gravity)-----------1 (t is time when they meet)
for q
s=x=30t -1/2gt^2 (u=30 and -1/2gt2 because motion opposite to gravity)----2
substituting the value of x in 2 to 1
u will get the t