Final answer:
The speed at which the cylinder will hit the ground is approximately 63.25 m/s. However, from the given options, the closest correct answer is 60 m/s (Option C).
Step-by-step explanation:
To calculate the speed at which the cylinder will hit the ground when dropped from a helicopter at a height of 200 m, we can use the formula derived from the equations of motion for uniformly accelerated motion (in this case, the acceleration due to gravity):
v² = u² + 2as
Where:
- v is the final velocity (the speed we want to find),
- u is the initial velocity (0 m/s since the cylinder is dropped),
- a is the acceleration due to gravity (10 m/s²),
- s is the distance (height from which the cylinder is dropped, in this case, 200 m).
Substituting the known values into the equation, we get:
v² = 0² + 2(10 m/s²)(200 m)
v² = 4000 m²/s²
v = √(4000 m²/s²)
v = 63.25 m/s
The final velocity v is approximately 63.25 m/s, but since we're asked to choose from given options (A, B, C, D), the closest correct answer is C) 60 m/s.