Question

A car starts from rest and accelerates at $5\, m / s ^{2}$ 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 $\left. g =10\, m / s ^{2}\right)$

• A. $20\, m / s , 5\, m / s ^{2}$
• B. $20\, m / s , 0$
• C. $20\, \sqrt{2}\, m / s , 0$
• D. $20\, \sqrt{2}\, m / s , 10\, m / s ^{2}$

Answer 1

The Correct Option is D

To determine the velocity and acceleration of the ball at \(t=6\, s\), we first need to analyze the motion of the car and the ball. The car is moving with constant acceleration and a person drops a ball from the car after 4 seconds. Let's solve this step by step:

1. Finding the velocity of the car at \( t = 4 \, s \):
   The car starts from rest and accelerates at \(5 \, m/s^2\). The velocity of the car at any time \( t \) can be calculated using the equation: \(v = u + at\), where \( u = 0 \) (initial velocity), \( a = 5 \, m/s^2 \), and \( t = 4 \, s \). \(v = 0 + 5 \times 4 = 20 \, m/s\).
2. Velocity of the ball when dropped:
   At \( t = 4 \, s \), the ball is dropped. It inherits the velocity of the car at that instant. Thus, the initial horizontal velocity of the ball is \(20 \, m/s\).
3. Finding the velocity of the ball at \( t = 6 \, s \):
   After being dropped, the ball is only affected by gravity in the vertical direction, with \( g = 10 \, m/s^2 \). We will calculate the vertical component of the velocity after 2 seconds (from \( t = 4 \, s \) to \( t = 6 \, s \)). 
   The vertical velocity at \( t = 6 \, s \) is: \(v_y = 0 + 10 \times 2 = 20 \, m/s\). 
   Thus, the resultant velocity of the ball is given by: \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 20^2} = 20 \sqrt{2} \, m/s\).
4. Acceleration of the ball:
   After being dropped, the ball's only acceleration is due to gravity, which is \( 10 \, m/s^2 \) downwards.

Therefore, the velocity and acceleration of the ball at \( t=6 \, s \) are \(20 \sqrt{2} \, m/s\) and \(10 \, m/s^2\), respectively.