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 $t = 18 \, s$. What is the value of $v$? $(Take\, g = 10\, m/s^2)$

• A. $75\,m/s$
• B. $55\,m/s$
• C. $40\,m/s$
• D. $60\,m/s$

Answer 1

The Correct Option is A

To find the initial velocity \( v \) of the second ball, let's analyze the situation step-by-step:

1. The first ball is dropped from rest, so its initial velocity \( u_1 = 0 \). Using the equation of motion: s_1 = u_1t_1 + \frac{1}{2}gt_1^2 = 0 \cdot 18 + \frac{1}{2} \cdot 10 \cdot 18^2 = 0 + 5 \cdot 324 = 1620 \, \text{m}
2. The second ball is thrown after 6 seconds. Hence, it travels for only 12 seconds until both balls meet. For the second ball, the equation of motion is: s_2 = u_2t_2 + \frac{1}{2}gt_2^2 Since \( t_2 = 12 \, s \) and initial velocity is \( v \), the equation becomes: s_2 = v \cdot 12 + \frac{1}{2} \cdot 10 \cdot 12^2 = 12v + 5 \cdot 144 = 12v + 720
3. Since both balls meet at 1620 meters, we equate \( s_1 \) and \( s_2 \): 1620 = 12v + 720 Simplifying, we get: 12v = 1620 - 720 = 900 v = \frac{900}{12} = 75 \, \text{m/s}

Therefore, the initial velocity \( v \) of the second ball is 75\,m/s. The correct answer is 75\,m/s.