Answer:
1.
A. Vf = 29.4 m/s
B. h = 44.1 m
2.
A. Vf = 24.5 m/s
B. h = 30.625 m
3.
A. Vf = 5.5 m/s
4.
A. t = 3.03 s
B. Vf = 29.7 m/s
Step-by-step explanation:
1.
A.
using 1st equation of motion:
Vf = Vi + gt
where,
Vf = Final Velocity = ?
Vi = Initial Velocity = 0 m/s (since, book starts from rest)
g = 9.8 m/s²
t = time = 3 s
Therefore,
Vf = 0 m/s + (9.8 m/s²)(3 s)
Vf = 29.4 m/s
B.
using 2nd equation of motion:
h = Vi t + (1/2)gt²
h = (0 m/s)(3 s) + (1/2)(9.8 m/s²)(3 s)²
h = 44.1 m
2.
A.
using 1st equation of motion:
Vf = Vi + gt
where,
Vf = Final Velocity = ?
Vi = Initial Velocity = 0 m/s (since, book starts from rest)
g = 9.8 m/s²
t = time = 2.5 s
Therefore,
Vf = 0 m/s + (9.8 m/s²)(2.5 s)
Vf = 24.5 m/s
B.
using 2nd equation of motion:
h = Vi t + (1/2)gt²
h = (0 m/s)(2.5 s) + (1/2)(9.8 m/s²)(2.5 s)²
h = 30.625 m
3.
A.
using 3rd equation of motion:
2gh = Vf² - Vi²
where,
Vf = Final Velocity = ?
Vi = Initial Velocity = 3 m/s
g = 9.8 m/s²
h = height = 2 m
Therefore,
2(9.8 m/s²)(2 m) = Vf² - (3 ms)²
Vf² = 39.2 m²/s² - 9 m²/s²
Vf = √30.2 m²/s²
Vf = 5.5 m/s
4.
A.
using 2nd equation of motion:
h = Vi t + (1/2)gt²
where,
h = height = 45 m
Vi = Initial Velocity = 0 m/s (since, photo copier was initially at rest)
t = time = ?
g = 9.8 m/s²
45 m = (0 m/s)(t) + (1/2)(9.8 m/s²)(t)²
t² = 2(45 m)/(9.8 m/s²)
t = √9.18 s²
t = 3.03 s
B.
using 1st equation of motion:
Vf = Vi + gt
Vf = 0 m/s + (9.8 m/s²)(3.03 s)
Vf = 29.7 m/s