SATHEE: Kinematics Question 408

Kinematics Question 408

Question: 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? $(t a k e g = 10 m / s^{2})$

Options:

A) 75 m/s

B) 55 m/s

C) 40 m/s

D) 60 m/s

Show Answer

Answer:

Correct Answer: A

Solution:

Clearly distance moved by 1st ball in 18s = distance moved by 2nd ball in 12s.

Now, distance moved in 18 s by 1st ball $= \frac{1}{2} \times 10 \times 18^{2} = 90 \times 18 = 1620 m$

distance moved in 12 s by 2nd ball $= u t + \frac{1}{2} g t^{2} ∴ 1620=12v+5 \times 144$
$\Rightarrow v=135-60=75 m s^{- 1}$

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Kinematics Question 408

Question: 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? (takeg=10m/s2)

Options:

Answer:

Solution:

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Question: 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? $(t a k e g = 10 m / s^{2})$