A car starts from rest and accelerates at 5 m/s^². At t = 4s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t = 6s? (Take g = 10 m/s^²)

Correct option is B

​$$\textbf{Given:} \\\bullet\ \text{Car acceleration: } a_{\text{car}} = 5\ \text{m/s}^2 \\\bullet\ \text{Ball is dropped at } t = 4\ \text{s} \\\bullet\ \text{Gravity } g = 10\ \text{m/s}^2 \\\bullet\ \text{Find velocity and acceleration of the ball at } t = 6\ \text{s} \\[10pt]\textbf{Step 1: Velocity of the car at the time ball is dropped} \\[4pt]\text{At } t = 4\ \text{s, the car's velocity is:} \\v_{\text{car}} = u + at = 0 + 5 \times 4 = 20\ \text{m/s} \\[4pt]\text{So, when the ball is dropped, it has:} \\\bullet\ \text{Horizontal velocity } = 20\ \text{m/s} \\\bullet\ \text{Vertical velocity } = 0\ \text{m/s} \\[10pt]\textbf{Step 2: Ball’s motion after being dropped (from } t = 4\ \text{to } t = 6\ \text{s)} \\[4pt]\bullet\ \text{Time since drop } = 6 - 4 = 2\ \text{s} \\\bullet\ \text{Ball undergoes projectile motion:} \\\quad \circ\ \text{Horizontal motion: constant velocity } v_x = 20\ \text{m/s} \\\quad \circ\ \text{Vertical motion: acceleration } g = 10\ \text{m/s}^2 \\[6pt]\text{Vertical velocity after 2 s:} \\v_y = 0 + gt = 10 \times 2 = 20\ \text{m/s}$$

$$\textbf{Step 3: Final velocity of the ball at } t = 6\ \text{s} \\[4pt]\text{It has two components:} \\\bullet\ \text{Horizontal: } v_x = 20\ \text{m/s} \\\bullet\ \text{Vertical: } v_y = 20\ \text{m/s} \\[6pt]v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 20^2} = \sqrt{800} = 20\sqrt{2}\ \text{m/s} \\[12pt]\textbf{Step 4: Acceleration of the ball} \\\bullet\ \text{Horizontal acceleration: } 0 \\\bullet\ \text{Vertical acceleration: } g = 10\ \text{m/s}^2 \\[6pt]\text{So,} \quad \boxed{\text{Acceleration of ball} = 10\ \text{m/s}^2 \text{ downward}}$$​​​