1. A curious physics student drops a book from a high building. A. What is the VELOCITY of the brick after 3.0s?
B. How FAR(Distance) does the book fall during this time?

2. Suppose a free-fall ride at an amusement park starts at a rest and then drops.
A. What is the VELOCITY of the ride after 2.5s?
B. How FAR do people on the ride fall during the 2.5s time period?

3. You throw a ball downward from a window at a speed of 3.0 m/s.
A. How FAST will it be moving when it hits the sidewalk 2.0m below?

4. If I drop a photo copier off of a 45m tall building and splats on the ground below:
A. Calculate the TIME to hit the ground:
B. Calculate the VELOCITY just before:

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

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .