Final answer:
The final velocity of an object falling from a 75m height is 38 m/s, calculated using the kinematic equations considering the acceleration due to gravity. The distance traveled during the first second is 4.9 meters using the equation for distance given constant acceleration.
Step-by-step explanation:
To determine the final velocity of an object that falls from a height of 75m, we can use the kinematic equations for an object in free fall. When an object is dropped from rest, its initial velocity (vo) is 0 m/s, and assuming the acceleration due to gravity (a) is approximately 9.8 m/s2, we have:
v2 = vo2 + 2 * a * d
Where v is the final velocity, d is the distance fallen (75m), and vo is the initial velocity. Plugging in the values:
v2 = 0 + 2 * 9.8 m/s2 * 75m
v2 = 1470 m2/s2
v = √1470 m2/s2 ≈ 38.3 m/s
The negative sign in option 2 would indicate a direction towards the ground, but since the question is looking for magnitude, the correct answer is 38 m/s (option 1).
Answering part (a) from the references, the distance traveled during the first second is found with d = vo * t + 0.5 * a * t2, where t is time in seconds. Since the initial velocity is zero, it simplifies to:
d = 0.5 * 9.8 m/s2 * (1 s)2
d = 4.9 m
So, the distance traveled during the first second is 4.9 meters. Regarding part (c), the distance traveled during the last second would require knowing the total time taken to hit the ground and then calculating the positions at the last and penultimate seconds which is beyond the scope of this particular answer.