Question

0. A ball is dropped from a high rise platform at t = 0
starting from rest. After 6 seconds another ball is
thrown downwards from the same platform with a
speed v. The two balls meet at t = 18 s. What is
the value of v? (Take g = 10 m/s2)
(1) 60 m/s
(2) 75 m/s
(2) 55 m/s
(4) 40 m/s
time t according​

Answer 1

Answer:

75 m/s

Explanation:

For first body:

Since it is dropped(not thrown with velocity), its initial velocity is 0.

u = 0, a = g, final velocity(at t = 18) be w.

Using S = ut + ½ at² at t = 18 for first ball:

=> S = 0(18) + ½ (10)(18)²

= 1620

Similarly for other ball but since it is thrown with velocity, initial velocity is v, with all same conditions as that of first one, excluding time.

As it is dropped after 12 sec, time taken to reach the same distance 1620 m should be t = 18 - 6 = 12 . Here,

=> 1620 = v(12) + ½ (10)(12)²

=> 1620 = 12v + 720

=> 75 = v