Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t =0 s. Ball B is thrown vertically down with an initial velocity ' u ' at t =2 s. After a certain time, both balls meet 100 m above the ground. Find the value of 'u' in ms -1. [use g =10 ms -2 ] :

Question

Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t =0 s. Ball B is thrown vertically down with an initial velocity ' u ' at t =2 s. After a certain time, both balls meet 100 m above the ground. Find the value of 'u' in ms -1. [use g =10 ms -2 ] : (A) 10 (B) 15 (C) 20 (D) 30

Tardigrade Admin · Accepted Answer

Let they meet at time t. t=√(2 h/g)=√(2 × 80/10) =4 sec Time taken by ball B to meet A =2 sec using S=u t+(1/2) at 2 -80=- u × 2+(1/2)(-10)(2)2 u =30

Q. Two balls A and B are placed at the top of 180m tall tower. Ball A is released from the top at t=0s. Ball B is thrown vertically down with an initial velocity ' u ' at t=2s. After a certain time, both balls meet 100m above the ground. Find the value of 'u' in ms−1. [use g=10ms−2 ] :

10

15

20

30

Let they meet at time t. t=g2h​​=102×80​​ =4sec Time taken by ball B to meet A=2sec using S=ut+21​at2 −80=−u×2+21​(−10)(2)2 u=30

Let they meet at time $t$ .
$t = \frac{2 h}{g} = \frac{2 \times 80}{10}$
$= 4 sec$
Time taken by ball $B$ to meet $A = 2 sec$
using $S = u t + \frac{1}{2} a t^{2}$
$- 80 = - u \times 2 + \frac{1}{2} (- 10) (2)^{2}$
$u = 30$