Question

1. A curious physics student drops a book from a high building.

Answer 1

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  

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