Answer
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Hint: This question utilises the concept of relative motion and free fall under gravity. Since the food packet is “dropped” from the helicopter, it does not have any initial velocity of its own. Its initial velocity is only relative to the helicopter which is rising against the direction of motion of the packet.
Formulae used:
$v = u + gt$
where $g$ is the acceleration due to gravity, $u$ is the initial velocity of the food packet, $t$ is the time passed and $v$ is the final velocity of the food packet.
Complete step by step solution:
To find the initial velocity of the food packet, we look at it with respect to the helicopter. Right before the fall, the helicopter was rising up with a velocity of $4m{s^{ - 1}}$ . This implies that the food packet was also rising up with a velocity of $4m{s^{ - 1}}$.
For sake of simplicity, we can assume the downward direction to be positive. Therefore, the available data is;
$g = + 9.8m{s^{ - 2}}$ where $g$ is the acceleration due to gravity.
$u = - 4m{s^{ - 1}}$ where $u$ is the initial velocity of the food packet.
$t = 3s$ where $t$ is the time passed
$v$ is the final velocity of the food packet. Now the first equation of free-fall motion is;
$v = u + gt$
where $g$ is the acceleration due to gravity, $u$ is the initial velocity of the food packet, $t$ is the time passed and $v$ is the final velocity of the food packet.
\[ \Rightarrow v = - 4 + 9.8 \times 3\]
\[ \Rightarrow v = - 4 + 9.8 \times 3\]
\[ \Rightarrow v = - 4 + 29.4\]
\[ \Rightarrow v = 25.4m{s^{ - 1}}\]
Therefore the velocity of the food packet after $3s$ is \[(B), 25.4m{s^{ - 1}}\].
Note: In such questions, it is important to check the direction of the initial velocity to avoid calculation mistakes. When an object is dropped from anywhere, by default it is not given any initial velocity of its own. In such cases, the initial velocity is due to inertia of motion.
Formulae used:
$v = u + gt$
where $g$ is the acceleration due to gravity, $u$ is the initial velocity of the food packet, $t$ is the time passed and $v$ is the final velocity of the food packet.
Complete step by step solution:
To find the initial velocity of the food packet, we look at it with respect to the helicopter. Right before the fall, the helicopter was rising up with a velocity of $4m{s^{ - 1}}$ . This implies that the food packet was also rising up with a velocity of $4m{s^{ - 1}}$.
For sake of simplicity, we can assume the downward direction to be positive. Therefore, the available data is;
$g = + 9.8m{s^{ - 2}}$ where $g$ is the acceleration due to gravity.
$u = - 4m{s^{ - 1}}$ where $u$ is the initial velocity of the food packet.
$t = 3s$ where $t$ is the time passed
$v$ is the final velocity of the food packet. Now the first equation of free-fall motion is;
$v = u + gt$
where $g$ is the acceleration due to gravity, $u$ is the initial velocity of the food packet, $t$ is the time passed and $v$ is the final velocity of the food packet.
\[ \Rightarrow v = - 4 + 9.8 \times 3\]
\[ \Rightarrow v = - 4 + 9.8 \times 3\]
\[ \Rightarrow v = - 4 + 29.4\]
\[ \Rightarrow v = 25.4m{s^{ - 1}}\]
Therefore the velocity of the food packet after $3s$ is \[(B), 25.4m{s^{ - 1}}\].
Note: In such questions, it is important to check the direction of the initial velocity to avoid calculation mistakes. When an object is dropped from anywhere, by default it is not given any initial velocity of its own. In such cases, the initial velocity is due to inertia of motion.
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