Final answer:
When the ball released from the helicopter hits the ground, it has both a horizontal velocity (unchanged from the helicopter) and a vertical velocity gained from falling under gravity. These are combined using the Pythagorean theorem to find the final velocity is approximately 59.7 m/s, so the closest option and the answer is 60 m/s.
Step-by-step explanation:
The student is asking about the final velocity of a ball that is released from a helicopter which is flying horizontally at 40 m/s at an altitude of 100 m. To solve this problem, we need to use the principles of physics, specifically the equations of motion for objects under gravity. The situation described involves two components of motion - horizontal and vertical. However, the horizontal velocity of the ball doesn't affect its vertical motion because these two components are independent of each other.
We know the initial vertical velocity (u) is 0 m/s since the ball is simply released (not thrown downwards or upwards), the gravitational acceleration (g) is approximately 9.81 m/s2, and the vertical distance (h) is 100 m. We ignore air resistance as per the instructions given.
To calculate the vertical velocity the ball will have when it hits the ground, we use the following equation:
v2 = u2 + 2gh,
where v is the final vertical velocity.
By plugging in the known values, we get
v2 = 0 + 2 * 9.81 m/s2 * 100 m,
v2 = 1962 m2/s2,
v = √(1962 m2/s2) = 44.3 m/s.
However, it will also have a horizontal component of velocity which is 40 m/s (since it maintains the velocity of the helicopter). The final velocity of the ball as it hits the ground is the vector sum of the horizontal and vertical components.
Using the Pythagorean theorem:
Final velocity (vf) = √(402 + 44.32) m/s,
Final velocity (vf) = √(1600 + 1962.49) m/s,
Final velocity (vf) = √(3562.49) m/s,
Final velocity (vf) ≈ 59.7 m/s.
Therefore, the correct option is 60 m/s when rounded to the nearest 10 m/s, which is a common rounding for answer choices.