A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 s another ball is thrown down

Question

A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 s 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 ms⁻²)

Accepted Answer

1.  **Motion of the First Ball (Dropped):**
    *   Initial velocity $u_1 = 0$.
    *   Time taken to meet $t_1 = 18$ s.
    *   Acceleration $a = g = 10$ ms⁻².
    *   Using the kinematic equation for displacement $h = ut + \frac{1}{2}gt^2$ :
        $$ h = 0 	imes 18 + \frac{1}{2} 	imes 10 	imes (18)^2 $$
        $$ h = 5 	imes 324 = 1620 	ext{ m} $$

2.  **Motion of the Second Ball (Thrown):**
    *   It is thrown 6 seconds later, so its time of flight is $t_2 = 18 - 6 = 12$ s.
    *   Initial velocity $u_2 = v$.
    *   It covers the same distance $h = 1620$ m.
    *   Applying the displacement equation:
        $$ 1620 = v(12) + \frac{1}{2} 	imes 10 	imes (12)^2 $$
        $$ 1620 = 12v + 5 	imes 144 $$
        $$ 1620 = 12v + 720 $$
        $$ 12v = 1620 - 720 = 900 $$
        $$ v = \frac{900}{12} = 75 	ext{ ms}^{-1} $$

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