Final answer:
After converting the packet's damage threshold to m/s and applying kinematic equations to analyze its fall, it is found that the packet hits the ground at approximately 21.9 m/s, exceeding the damage threshold of 20 m/s. Hence, the packet was likely damaged upon impact.
Step-by-step explanation:
To answer whether the packet dropped from the helicopter was damaged, we first need to convert the damage threshold velocity from km/hr to m/s. We have 72 km/hr which is equivalent to 20 m/s (since 1 km/hr = 0.27778 m/s). Now let's analyze the packet's impact velocity.
The motion of the packet can be described by the equations of motion under constant acceleration due to gravity. The packet was released from a height of 24 m with an initial upward velocity of 2 m/s. Using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration (9.8 m/s2 for gravity), and t is the time, we can find the time at which the packet comes to an instantaneous rest (its upwards velocity is zero).
For the upwards motion where v = 0 m/s (at the highest point), we have:
0 = 2 m/s + (-9.8 m/s2)t (since the acceleration acts downwards while the packet is moving upwards)
Solving for t gives us roughly 0.204 seconds.
The packet will now begin its descent from its highest point, which we can find by using the formula s = ut + 1/2at2, where s is the distance. Inserting the values, we get a height of approximately 24.204 m. From this point, the packet now falls under gravity from rest. We can find the impact velocity using the formula v^2 = u^2 + 2as. Since the initial descent velocity u is 0, we calculate:
v2 = 0 + 2(9.8 m/s2)(24.204 m)
Solving for v gives us a velocity of roughly 21.9 m/s upon impact.
Since 21.9 m/s is greater than 20 m/s (the damage threshold), the packet was likely damaged upon impact with the ground. Therefore, the answer is (a) Yes, the packet was damaged.