Two towers A and B , each of height 20 ~m are situated a distance 200 ~m apart. A body thrown horizontally from the top of the tower A with a velocity 20 ~ms ⁻¹ towards the tower B hits the ground at point P and another body thrown horizontally from the top of tower B with a velocity 30 ~ms ⁻¹ towards the tower A hits the ground at point Q . If a car starting from rest from P reaches Q in 10 seconds, then the acceler

Step-by-step solution

Given, height of both towers is same, h₁=h₂=h . Time of flight, t= 2 h g will be same t= 2 20 10 = 4 =2 ~s Displacement in horizontal direction from tower A to point P=u_A t =20 2=40 ~m Displacement in horizontal direction from tower B to point Q=u_B t =30 2=60 ~m So, distance between point P and Q aligned & =200-(40+60) & =100 ~m aligned Given, distance between P and Q is covered by car in 10 ~s , so using aligned s= & u t+ 1 2 a t^2 100= & 0 10+ 1 2 a(10)^2 & [ u=0, as car starts from rest ] a= & 2 aligned Accele