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)

First find the velocity of the car at the instant the ball is released ( t = 4 s ). The car starts from rest and accelerates uniformly at 5 m/s², so

$v\_\{car\}=a\backslash ,t\; =\; 5\; \backslash times\; 4\; =\; 20\; \backslash text\{ m/s\}$

This velocity is purely horizontal. When the ball is dropped it retains this horizontal component, while its vertical component is initially zero.

After release the ball is acted on only by gravity ( g = 10 m/s² ) for the next 2 seconds (from t = 4 s to t = 6 s). The vertical velocity after Δt = 2 s is

$v\_\{y\}=g\backslash ,\backslash Delta\; t\; =\; 10\; \backslash times\; 2\; =\; 20\; \backslash text\{ m/s\; (downward)\}$

The horizontal velocity remains unchanged:

$v\_\{x\}=20\; \backslash text\{ m/s\; (horizontal)\}$

Combine the two orthogonal components to obtain the speed of the ball at t = 6 s:

$v\; =\; \backslash sqrt\{v\_\{x\}^\{2\}+v\_\{y\}^\{2\}\}\; =\; \backslash sqrt\{20^\{2\}+20^\{2\}\}\; =\; 20\backslash sqrt\{2\}\backslash \; \backslash text\{m/s\}$

The only acceleration acting on the ball after it is released is gravity, giving a constant vertical acceleration of 10 m/s² (no horizontal acceleration). Hence the magnitude of the ball’s acceleration is

$a\; =\; 10\; \backslash text\{ m/s\}^\{2\}$

Therefore, at t = 6 s the ball’s velocity is 20√2 m/s and its acceleration is 10 m/s².