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Step-by-Step Solution
Step 1: Determine the car’s velocity at t = 4 s
The car starts from rest and accelerates at
$a = 5\,\text{m/s}^2$.
Using the relation
$v = u + at$,
with
$u = 0$
and
$t = 4\,\text{s}$,
we get:
$v_{\text{car}} = 0 + (5\,\text{m/s}^2)(4\,\text{s}) = 20\,\text{m/s}$.
Step 2: Establish the ball’s initial conditions at the instant it is dropped
At $t = 4\,\text{s}$, the ball is dropped. Its initial horizontal velocity equals the car’s velocity at that moment:
$V_{x,\text{initial}} = 20\,\text{m/s}$.
Because the ball is simply released (not propelled downward or upward), its initial vertical velocity at the moment of release is:
$V_{y,\text{initial}} = 0\,\text{m/s}$.
Step 3: Calculate the ball’s horizontal velocity at t = 6 s
No horizontal acceleration acts on the ball (ignoring air resistance). Hence, its horizontal velocity remains constant at:
$V_{x} = 20\,\text{m/s}$.
Step 4: Calculate the ball’s vertical velocity at t = 6 s
From $t = 4\,\text{s}$ to $t = 6\,\text{s}$, the ball is in free fall for $2\,\text{s}$. The acceleration due to gravity is $g = 10\,\text{m/s}^2$ downward. The vertical velocity is given by
$v_{y} = V_{y,\text{initial}} + g \times \Delta t$,
where
$\Delta t = 2\,\text{s}$.
Since $V_{y,\text{initial}} = 0$:
$V_{y} = 0 + 10\,\text{m/s}^2 \times 2\,\text{s} = 20\,\text{m/s}$ (downward).
Step 5: Find the magnitude of the ball’s velocity at t = 6 s
At $t = 6\,\text{s}$, the ball’s velocity components are:
Horizontal: $V_{x} = 20\,\text{m/s}$
Vertical: $V_{y} = 20\,\text{m/s}$ (downward)
The resultant velocity is found by the Pythagorean formula:
$V = \sqrt{V_{x}^2 + V_{y}^2} = \sqrt{(20)^2 + (20)^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2}\,\text{m/s}.$
Step 6: Determine the ball’s acceleration at t = 6 s
After being released, the only acceleration on the ball is the acceleration due to gravity, which is $10\,\text{m/s}^2$ downward. Therefore:
$\text{Acceleration} = g = 10\,\text{m/s}^2 \text{ (downward)}.$
Answer
The velocity of the ball at $t = 6\,\text{s}$ is $20\sqrt{2}\,\text{m/s}$, and its acceleration at that instant is $10\,\text{m/s}^2$ downward.
