A ball is dropped from a high rise platform at t = 0 starting from rest

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 = 18s. What is the value of v?(take g = 10 m/s²)

Options

(a) 75 m/s
(b) 55 m/s
(c) 40 m/s
(d) 60 m/s

Correct Answer:

75 m/s

Explanation:

Clearly distance moved by 1st ball in 18 s = distance moved by 2nd ball in 12 s. Now, distance moved in 18 s by 1st ball 1/2 x 10 x 18²
= 90 x 18 = 1620 m. Distance moved in 12 s by 2nd ball = ut + 1/2 gt²
1620 = 12v + 5 x 144
v = 135 – 60 = 75 ms

admin:

Related Questions

  1. A body moves with a uniform acceleration a and zero initial velocity
  2. A point charge is kept at the centre of a metallic insulated spherical shell. Then
  3. A certain metallic surface is illuminated with monochromatic light of wavelength
  4. In Bohr model of hydrogen atom, the force on the electron depends on the principal
  5. Four point masses, each of value m, are placed at the corners of a square ABCD