1. Velocity of the truck at the instant of release

The truck starts from rest and moves with uniform acceleration $a_x = 5\,\text{m s}^{-2}.$

So after $20\,$s, its velocity is $v_x = u_x + a_x t = 0 + 5\times 20 = 100\,\text{m s}^{-1}.$

When the ball is released, it has the same horizontal velocity as the truck: $v_{x0} = 100\,\text{m s}^{-1}.$

2. Horizontal motion of the ball

After release, no horizontal acceleration acts on the ball (neglect air resistance), so horizontal velocity remains constant: $v_x = 100\,\text{m s}^{-1}.$

3. Vertical motion of the ball

At the moment of release, the ball has no vertical velocity: $v_{y0} = 0.$

It falls under gravity for $1\,$s with $a_y = -g = -10\,\text{m s}^{-2}.$

Thus vertical velocity at impact is $v_y = v_{y0} + a_y t = 0 + (-10)(1) = -10\,\text{m s}^{-1}.$

So, $\vec v = 100\hat i - 10\hat j\,\text{m s}^{-1}.$

4. Checking options

• A: $100\hat i - 10\hat j$ ✅
• B: $10\hat i - 100\hat j$ ❌
• C: $100\hat i$ ❌ (ignores vertical velocity)
• D: $-10\hat j$ ❌ (ignores horizontal velocity)

Therefore, the correct answer is Option A.