How Do You Solve Basic Gravity Problems in Physics?

[ihopeican]

Homework Statement

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

Homework Equations

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

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

 

[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

 

[ihopeican]

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[/QU

Hello.

I really took that first comment to heart :'(. all i relised i was doing is not putting a as a negative value. and i would have gone over the topic with my teacher to revise it but we hadnt done it yet and he just gave it to me to try to do.
anyway thankyou.

 

[ihopeican]

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

 

[learningphysics]

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

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

 

[ihopeican]

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

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.

 

[learningphysics]

Hello,
I got 11.47 for the first one (1.a).

Correct.

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.

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:

vf^2 = vi^2 + 2ad

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.

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

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.

For 2.a) i got 39.6

Correct.

For 2.b) I got -27.85

Incorrect. Can you show your calculations?