HELP ME! omg?! Physics again?

Question

Oh, please guys! Help me.

Here are the questions:

1. A stone is thrown upward with an initial speed of 20 min/s. It goes up to a certain height, then comes down to the same level from where it it was thrown, assume g=10m/s^2.

a. How high does the stone go before it starts to come down? How long does it take to the stone to reach this point?

b. What is the acceleration of the stone at the highest point it reached? What is its speed at the highest?

c. How long does it take the stone to reach the same level from were it was thrown?

Note: Express your answer in each letter; 4 significant figures.

Oh, please guys. HELP ME coz I'll submit it tomorrow.

love you. :*

Answer 1

a. 40 m

b. 10m/s^2

c. 4 seconds

idk, I don't really remember :T

Answer 2

Initial velocity u = 20 min/s

Acceleration a = -g = -10 m/s^2 (negative because stone is thrown upward and gravity is downward)

a)For maximum height, final velocity v = 0

v^2 = u^2 + 2as

Or, 0 = 20^2 - 2*10*s

Or, s = 20^2/(2*10) = 400/20 = 20 m

v = u + at

Or, 0 = 20 -10 * t

Or, t = 20/10 sec = 2 sec

Ans: Stone goes to the height of 20 m before it starts to come down. It takes 2 sec to reach this point.

b) Acceleration at the highest point is 10 m/s^2 vertically downward. Speed at the highest point is 0.

c) Stone takes 2 sec to go to the highest point. It takes further 2 sec to come from the highest point to the level it was thrown.

So, total time = 2 sec + 2 sec = 4 sec

Answer 3

Well you can use the formula v^2 = u^2 + 2as

U is initial speed and V is velocity. S is distance and a is accelerating due to gravity. When the stone reaches the top of it's flight (how high up it goes) it's velocity is 0 as it has stopped moving. Therefore we can fill in equation.

0= 20^2 + 2(10)s

You can then solve for s (do this urself i don't have a calculator on me) :)

b. Acceleration at highest point is -10 as gravity is a constant -- just velocity changes - velocity is 0 - as it stops moving to come back down

c. Use v=u+at to find out time

0=20+(10)t

-- Then Solve for t

As the acceleration is always constant it takes the same time for the stone to go up as to come back down (Just multiply the time found by two.)

Hope that helped  :)

Answer 4

You've got three formulae to choose from..

a) v = u + at

b) v² = u² + 2aS

c) S = ut + 0.5t².

For the first you have to use a)  to find the time as you don't know the distance.

u=20 v =0 a = -10

1) 0= 20-10t

10t = 20

t = 2 seconds to reach the top.

To find the distance you use b) v² = u² + 2 aS

0 =400 - 2x10xS  (- as you're throwing against gravity)

0=400-20S

20S =400

S = 400/20

S = 20 metres (to the top).

b) Acceleration is 10m/s² and is its speed is zero

c) It takes just as long to fall back down as it did to get up there so total time from your hand to your hand = 2+2 =4 secs

Answer 5

a. H=u^2/2g ; u=20

so H=20 and t=u/g so t=2sec

b. at the highest point the velocity becomes 0 and accelaration still remains -g= -10

c. 2t=4sec

Answer 6

Before you can answer these questions and ubderstand what is happening, there are some fundamental physics that you have to ubderstand:

1 When the stone thrown up wards it loses velocity because of the acceleration due to gravity acting downwards on the stone.

2. When the stone reaches its maximum height its velocity is zero, but the acceleraion of 10m/s² is still acting downwards.

Acceleration and displacement are vector quantities and so they have magnitude as well as direction. I find that most tudents have problems because they do not understand the importance of direction and the signs (+ or-) associated with direction. Time is a scalar quantity, it has no direction, only magnitude, and it is impossible to have a negative time as an answer.

There are 3 equations of motion which enable you to solve these problems.

I always advise my students to make a simple sketch of what is happening and to label it with the given information. Unfortunately it is not possible to draw a sketch here

For this problem we will take the direction UP as POSITIVE. DOWN is therefore NEGATIVE.

1) We will use the following equation of motion:

v² = u² +2*a*s

Where v = final velocity (in this case v=0)

u = initial velocity - in this case +20m/s note +ve sign because initial velocity is up

a -= acceleration due to gravity - in this case ( -10)m/s² Note -ve sign because acceleration is down.

Substitute:

0 = 20² +2*(-10) *s

0 = 400 - 20s

20s = 400

s = 20m

The stone will reach a height of 20m above its starting point

2. As already explained, acceleration at the highest point = 10m/s² downwards. The velocity at this point is =0  Note it is technically incorrect to say that the acceleratin is -10m/s².

3 The classic way to do this problem is as follows. There are many solutions which entail the calculation of the time to go up and the time to come doewn, but what I will do for you is the proper scientific use and understanding of the equations of motion:

For this problem use the equation

S = ut + ½ a*t²

Where

s = displacement , which is = 0 because the stone returned to its starting point

u = initial velocity, as above +20m/s

a = accel due to gravity, = -10m/s² because accel is downwards

t = time in seconds always +

Substitute:

0 = +20t + ½*(-10) t²

0 = 20t -5t²  

5t² -  20t = 0

( Factorise)

t(5t-20)=0

therefore t = 0 (ignore) or t = 4 seconds.

Answer time to return is 4 seconds.

Please note : If you did not pay special attention to signs in the last part of the problem, you would not solve the problem properly.

I do not follow the instruction to express the answers to 4 sinificant figures.