Question

A car starts from rest and accelerates at 5 m/s² . At t=4 s, 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=6 s? (Take g=10 m/s²)

• A. 20√2 m/s, 10 m/s²
• B. 20 m/s, 5 m/s²
• C. 20 m/s, 0
• D. 20√2 m/s, 0

Answer 1

The Correct Option is A

To solve the given problem, we need to determine the velocity and acceleration of the ball at \( t = 6 \, \text{s} \) after it is dropped from a window of the car.

1. First, let's consider the state of the car up to \( t = 4 \, \text{s} \). The car is accelerating from rest with an acceleration \( a = 5 \, \text{m/s}^2 \).
2. Using the equation of motion for velocity, \( v = u + at \) where \( u = 0 \, \text{m/s} \) (initial velocity), the velocity of the car at \( t = 4 \, \text{s} \) is:
   \(v = 0 + 5 \times 4 = 20 \, \text{m/s}\)
3. When the ball is dropped from the window at \( t = 4 \, \text{s} \), it will have the same horizontal velocity as the car at that moment, which is \( 20 \, \text{m/s} \). This becomes the horizontal component of the ball's velocity.
4. Vertically, the ball is subjected to gravitational acceleration \( g = 10 \, \text{m/s}^2 \) without any initial vertical velocity component. Thus, its vertical velocity at \( t = 6 \, \text{s} \) can be calculated using the formula:
   \(v_y = u_y + gt = 0 + 10 \times 2 = 20 \, \text{m/s}\) (considering the time interval after drop, \( \Delta t = 6 - 4 = 2 \, \text{s} \))
5. The resultant velocity of the ball at \( t = 6 \, \text{s} \) is given by the vector addition of horizontal and vertical components:
   \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 20^2} = \sqrt{800} = 20\sqrt{2} \, \text{m/s}\)
6. The acceleration of the ball at \( t = 6 \, \text{s} \) is due to gravity acting downward, which is \( 10 \, \text{m/s}^2 \).

Thus, the velocity of the ball at \( t = 6 \, \text{s} \) is \( 20\sqrt{2} \, \text{m/s} \) and its acceleration is \( 10 \, \text{m/s}^2 \). Therefore, the correct option is 20√2 m/s, 10 m/s².