Motion problem when acceleration is constant

[brsullivan]

Homework Statement

A coin, initially at rest is dropped from an elevated position and freely falls to the ground with an acceleration of 10m/s/s.
a) How fast is the coin traveling after 2 sec b) How fast is the coin traveling after 5 seconds
c) How far does the coin fall after 2 sec d) How far does the coin fall after 5 sec
e) How much time does it take the coin to fall 5m f) How much time does it take the coin to fall 10m g) What is the velocity of the coin after it has fallen 5 m h) What is the velocity of the coin after it has traveled 10m i) A feather is dropped at the same time as the coin and falls with an acceleration of 3.0 m/s/s (due to air resistance). What is the speed of the feather after the coin has fallen 5m

Homework Equations

I don't know

The Attempt at a Solution

I am terrible at physics and have absolutely no idea what to do. I don't even know where to start. I am a high school student in college physiscs...so yeah.

 

[rl.bhat]

1.

Homework Equations

I don't know

.

Go through any physics textbook and find out the relevant equations.

 

[baba89]

I can provide guidance on how to approach this problem and solve it using the relevant equations and principles. First, let's define the given variables:

- Initial velocity (u) = 0 m/s (since the coin is initially at rest)
- Acceleration (a) = 10 m/s^2 (given in the problem statement)
- Time (t) = 2 sec and 5 sec (given in parts a and b)

a) To find the velocity of the coin after 2 seconds, we can use the formula v = u + at, where v is the final velocity. Plugging in the given values, we get:

v = 0 + (10 m/s^2)(2 sec) = 20 m/s

Therefore, after 2 seconds, the coin is traveling at a speed of 20 m/s.

b) Similarly, to find the velocity after 5 seconds, we use the same formula:

v = 0 + (10 m/s^2)(5 sec) = 50 m/s

c) To find the distance traveled by the coin after 2 seconds, we can use the formula s = ut + 1/2at^2, where s is the distance traveled. Plugging in the values, we get:

s = (0 m/s)(2 sec) + 1/2 (10 m/s^2)(2 sec)^2 = 20 m

Therefore, after 2 seconds, the coin has fallen 20 meters.

d) Similarly, to find the distance traveled after 5 seconds, we use the same formula:

s = (0 m/s)(5 sec) + 1/2 (10 m/s^2)(5 sec)^2 = 125 m

e) To find the time taken for the coin to fall 5 meters, we can rearrange the formula s = ut + 1/2at^2 to solve for time:

t = √(2s/a)

Plugging in the values, we get:

t = √(2(5 m)/(10 m/s^2)) = 1 sec

Therefore, it takes the coin 1 second to fall 5 meters.

f) Similarly, to find the time taken for the coin to fall 10 meters, we use the same formula:

t = √(2(10 m)/(10 m/s^2)) = √2 sec