# Homework Help: Basic Gravity Worksheet help.

1. Sep 22, 2007

### ihopeican

1. The problem statement, all variables and given/known data
1.) Astone is thrown upwards with a speed of 15ms-1
a.) how high does it reach
b.)with what speed does it hit the ground
c.)how long does it spend in flight

2. Relevant equations
2as=ut+1/2at^2
v=s/t

3. The attempt at a solution

Well i am not sure what initial or final velocity is all i can figure out is that the acceleration is 9.8ms-2

2.) A ball is thrown down from the top of a cliff at 10ms-1. If it taks 2 seconds to reach the bottom of the cliff, Calculate..
a.) how high is the cliff.
b.)with what speed does it hit the bottom.

Again i can't identify the information into V,U, etc.

I am basically having trouble identifying u,v,t,a etc.

Thankyou

2. Sep 22, 2007

### Vidatu

This is really, really basic stuff. If you have problems with this, you'll have more problems later. Take some time and meet with your instructor to go over these topics again.

Anyway;

For question 1, I'll assume that we don't know how high the stone is when released, so we can ignore that. The only force it is acting against is gravity, which you have identified as 9.8m/s^2 (remember that it is a negative value). Your initial velocity is what speed it is thrown at, which is given in the question. It is easiest to say that final velocity is 0 (when the stone is at the top of its arc).
Use the formula (Vf)^2 = (Vi)^2 + (2)(a)(d) for parts a and b
Use t = 2 ( (2d)/(Vf-Vi) ) for part c

3. Sep 23, 2007

### ihopeican

4. Sep 23, 2007

### ihopeican

can anyone confirm some answers for me so i know if im right for questions 1,abc and question 2

5. Sep 23, 2007

### learningphysics

sure... post your calculations and answers... it will make it easy to confirm and to point out any errors.

6. Sep 23, 2007

### ihopeican

Hello,
I got 11.47 for the first one (1.a).
I got 4.5 for 1b but i don't think this one is right. i did 2x-9.8x11.25 and then did the square root of that.

For 1.c i am not sure with this one so some help would be much appreciated.

For 2.a) i got 39.6

For 2.b) I got -27.85

Thankyou lots of mistakes i bet but i am just not thinking straight today and we only just started our physics unit.

7. Sep 23, 2007

### learningphysics

Correct.

By conservation of energy, you should get back the original speed 15m/s. But the velocity is the opposite direciton... so velocity is 15m/s downward. The kinematics equations will give you back that answer... try using:

[(vi + vf)/2]*t = d

d = 0

so

[(vi + vf)/2]*t = 0

vi + vf = 0

vf = -vi

vf = -15m/s

You can also use:

you want d = 0... you get vf^2 = vi^2... so vf = +-vi. You know it is coming back down... so taking up as positive and down as negative... vf = -15m/s.

Two methods come to mind... Find the time to reach maximum height, multiply by 2... that's the total time of flight...

Or try using

d = vt*t + (1/2)(-g)t^2

d = 0.

solve for t.

Correct.

Incorrect. Can you show your calculations?