Engineering Science

[匿名]

2009-10-15 12:11 am

Take acceleration due to gravity g=10m/s2
2. You are standing at the foot of the Bank of America building in San Francisco, which is
52 floors (237 m) high. You launch a ball straight up in the air from the edge of the foot
of the building. The initial vertical speed is 70 m/s. (For this problem, you may ignore
your own height, which is very small compared to the height of the building.)
a. How high up does the ball go?
b. How fast is the ball going right before it hits the top of the building?
c. For how many seconds total is the ball in the air?

3. You drop a rock from the top of a cliff. The rock takes 3.5
seconds to reach the bottom.
a. What is the initial speed of the rock?
b. What is the magnitude (i.e. numerical value) of the
acceleration of the rock at the moment it is dropped?
c. What is the magnitude of the acceleration of the rock
when it is half-way down the cliff?
d. What is the height of the cliff?

4. Draw free body diagrams (FBDs) for all of the following objects involved (in bold) and
label all the forces appropriately. Make sure the lengths of the vectors in your FBDs are
proportional to the strength of the force: smaller forces get shorter arrows!
a. A man stands in an elevator that is accelerating upward at 2 m/s2.
b. A boy is dragging a sled at a constant speed. The boy is pulling the sled
with a rope at a 30° angle.
c. The picture shown here is attached to the ceiling by three wires.

回答 (1)

Prof. Physics

2009-10-16 5:40 am

✔ 最佳答案

2.a. By v^2 = u^2 - 2gs

0 = (70)^2 - 2(10)s

Maximum height of the ball from the foot of the building, s = 245 m

b. By v^2 = u^2 - 2gs

v^2 = (70)^2 - 2(10)(237)

Speed, v = 12.6 m/s

c. By v = u - gt

-12.6 = 70 - (10)t (Assume the ball hits the top of the building when it is on its way downward.)

Time of flight, t = 8.26 s

3.a. Initial speed of the rock = 0

b. Acceleration is due to gravity, so the numerical value is 10 ms^-2

c. The acceleration is the same throughout the journey, so it is still 10 ms^-2.

d. By s = ut + 1/2 gt^2

s = 0 + 1/2 (10)(3.5)^2

Height, s = 61.25 m

4.a. I will not draw the Free body diagram, but I can give you the guideline.

There are two forces acting on the man, namely the reaction on him by the floor of the elevator, which acts upward, and his own weight, acting downward.

The reaction should be larger than his weight, as he is accelerating upwards.

b. There are four forces acting on the sled.

The tension in the rope, making an angle 30* with the horizontal.

Vertically, the reaction, and the weight. The weight should be larger than the weight.

Horizontally, the friction (since the sled is under constant velocity.) friction should be smaller than the tension in the rope.

參考: Physics king

👍 0 👎 0