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 this problem, let's break it down step-by-step. 

1. Understanding the Motion of the Car:

• The car accelerates from rest with a constant acceleration \(a = 5 \, \text{m/s}^2\).
• We need to find the velocity of the car (and hence the horizontal velocity of the ball at the instant it is dropped) at \(t = 4 \, \text{s}\).

The formula for velocity under constant acceleration is given by:

\(v = u + at\)

Here, \(u = 0 \, \text{m/s}\) (since the car starts from rest), \(a = 5 \, \text{m/s}^2\), and \(t = 4 \, \text{s}\). Substituting these values, we get:

\(v = 0 + 5 \times 4 = 20 \, \text{m/s}\)

This means at the moment the ball is dropped, it has a horizontal velocity of \(20 \, \text{m/s}\).

2. Understanding the Motion of the Ball After Being Dropped:

• The ball is subject only to gravitational acceleration in the vertical direction after being dropped.
• The horizontal velocity of the ball remains constant at \(20 \, \text{m/s}\) since there are no forces acting in the horizontal direction.
• The vertical acceleration of the ball is \(g = 10 \, \text{m/s}^2\) downward.

3. Calculating the Vertical Velocity of the Ball at \(t=6 \, \text{s}\):

• Since the ball is dropped at \(t = 4 \, \text{s}\), we are interested in its motion between \(t = 4 \, \text{s}\) and \(t = 6 \, \text{s}\).
• The time of vertical motion \(\Delta t = 2 \, \text{s}\).

The vertical velocity after a time \(\Delta t\) is given by:

\(v_y = 0 + gt = 10 \times 2 = 20 \, \text{m/s}\)

4. Resultant Velocity of the Ball at \(t=6 \, \text{s}\):

To find the magnitude of the resultant velocity, we use the Pythagorean theorem:

\(v_{\text{resultant}} = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 20^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \, \text{m/s}\)

5. Acceleration of the Ball at \(t=6 \, \text{s}\):

The acceleration of the ball is only due to gravity, which is \(10 \, \text{m/s}^2\) downward.

Thus, the velocity and acceleration of the ball at \(t = 6 \, \text{s}\) are 20√2 m/s and 10 m/s² respectively.