Question

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 ?
A)20m/s, 5m/s^2.
B)20m/s,0
C)20√2m/s,0.
D)20√2m/s,10m/s^2


Please help me.​

Answer 1

Answer:

right now I have a lot of money to make it to the table and everyone else will be there

Answer 2

Given:

Initial velocity of the car $\[ = 0\]$

Acceleration of the car $\[ = 5m/{s^2}\]$

Time at which the ball is dropped $t=4s$

To Find: The velocity and acceleration of the ball at $t=6s$

Solution:

From the equations of motion,

$\[v = u + at\]$

Therefore, the velocity of the car at $t=4s$ is

$\[v = 0 + 5m/{s^2} \times 4s\]$

$\[v = 20m/s\]$

As the ball is dropped from the window of the car, it moves with acceleration due to gravity. Therefore, the acceleration of the ball at $t=6s$ will be equal to the value of acceleration due to gravity, i.e. $\[9.8m/{s^2} \approx 10m/{s^2}\]$.

At $t=6s$, the velocity of the ball can be given as:

$\[v = \sqrt {{v_x}^2 + {v_y}^2} \]$

Here, $\[{{v_x}}\]=$ velocity due to car, i.e. $\[20m/s\]$ and,

$\[{{v_y}}\]=$ velocity of the ball moving downward

$\[{{v_y}}\]$ can be given as:

$\[{v_y} = u + gt\]$

$\[{v_y} = 0 + 10m/{s^2} \times 2s\]$   (here, $t=6s-4s=2s$)

$\[{v_y} = 20m/s\]$

Therefore, the velocity of the ball can be given as

$\[v = \sqrt {{{20}^2} + {{20}^2}} \]$

$\[v = 20\sqrt 2 m/s\]$

Hence, the velocity and acceleration of the ball at $t=6s$ are $\[20\sqrt 2 m/s\]$ and $\[10m/{s^2}\]$ respectively.

Thus, the correct option is D.