Free fall quiz-assume air resistance is ignored

1. luke drops a pile of roof shingles from the top of a roof located 8.52m above the ground. determine the time required for the shingles to reach the ground.
2. rex throws his mother’s crystal vase vertically upwards with an initial velocity of 26.2m/s. determine the height to which the vase will rise above the initial height. at what velocity is the vase moving when rex catches it.
3. an object falls from the transco tower in houston and takes 15seconds to reach the ground. what is the velocity at impact and how tall is the building?
4. an orange is tossed upward at 21m/s. what is the velocity of the orange 3.5s later? what is the height of the orange at this time? is the orange still travelling up or is it travelling down?
5. a tennis ball is tossed up off a building with a velocity of 22m/s. it takes 6.4s to reach the ground. how high is the building and what is the maximum height of the tennis ball?

Q1. The answer is 1.32 s.

The kinetic equation used to calculate the time is:

d = v1 t + 1/2 at²

d - distance

v1 - initial velocity

t - time

a - gravitational acceleration

We know:

d = - 8.52 m (negative because it is downwards)

v1 = 0      (there is no initial velocity in free fall)

a =  - 9.8 m/s² (negative in free falls)

d = v1 t + 1/2 at²

- 8.52 = 0 * t + 1/2 * - 9.8 * t²

- 8.52 = 0 - 4.9 t²

-8.52 = -4.9 t²

t² = -8.52/-4.9

t² = 1.74

t = √1.74

t = 1.32 s

Q2. PART A) The answer is 35 m.

The kinetic equation that can be used is:

v2² = v1² +2ad

d - distance

v1 - initial velocity

v2 - final velocity

a - gravitational acceleration

We know:

v2 = 0    (The final velocity is at the peak so it is 0)

v1 = 26.2 m/s

a = 9.8 m/s²   (negative in free falls)

v2² = v1² +2ad

0 = 26.2² + 2 * ( - 9.8) * d

0 = 686.44 - 19.6d

19.6d = 686.44

d = 686.44/19.6

d = 35.0 m

Q2. PART B) The answer is 26.2 m/s.

The velocity at what is the vase moving when Rex catches is v2. 

v1 - initial velocity

v2 - final velocity

a - gravitational acceleration

d - distance

We know:

v2 = ?

v1 = 0      (in free fall)

a = - 9.8 m/s² (negative in free falls)

v2² = v1² + 2*a*d

v2² = 0 + 2 * -9.8 * -35

v2² = 686

v2 = √686

v2 = 26.2 m/s

Q3. PART A) The answer is -147 m/s

The kinetic equation that can be used is:

v2 = v1 + a * t

v1 - initial velocity

v2 - final velocity

a - gravitational acceleration

t - time

We know:

v1 = 0      (the initial speed is 0 at free fall)

v2 = ?

a = - 9.8 m/s² (negative in free fall)

t = 15s

v2 = v1 + a * t

v2 = 0 + -9.8 * 15

v2 = -147 m/s

Q3. PART B) The answer is 1102.5 m.

The kinetic equation that can be used is:

v2² = v1² +2ad

d - displacement

v1 - initial velocity

v2 - final velocity

a - gravitational acceleration

We know:

v2 = -147 m/s

v1 = 0 (in free fall)

a = -9.8 m/s²   (negative in free falls)

The kinetic equation that can be used is:

v2 = v1 + a * t

v1 - initial velocity

v2 - final velocity

a - gravitational acceleration

t - time

We know:

v2 = ?

v1 = 21 m/s

a = 9.8 m/s (negative in free fall)

t = 3.5s

v2 = v1 + a * t

v2 = 21 + (-9.8)*3.5

v2 = 21 - 34.3

v2 = -13.3 m/s

Q4. PART B) The answer is 9.025 m

The kinetic equation that can be used is:

v2² = v1² +2ad

d - displacement

v1 - initial velocity

v2 - final velocity

a - gravitational acceleration

v2² = v1² +2ad

d - displacement

v1 - initial velocity

v2 - final velocity

a - gravitational acceleration

We know:

v2 = 0  (when it reaches the ground)

v1 = 22 m/s

t = 6.4 s

a = -9.8 m/s (in free fall)

The kinetic equation used to calculate the time is:

v2 = v1 + a*h

h - height

v1 - initial velocity

v2 - final velocity

a - gravitational acceleration

We know:

v1 = 22 m/s

v2 = 0

h = ?

a = -9.8 m/s (negative in free fall)

v2 = v1 + a*h

0 = 22 + -9.8 * h

-22 = -9.8h

h = -22/-9.8

h = 2.24 m

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.