gravitational acceleration

[BTICU]

I've been given a challenge by my grade 11 physics teacher and I can't make heads of tail of it, I’ve got the formulas but the numbers don't make sense.

 

If you went up 109 km (above sea level) and dropped a rock, what speed would it be traveling at and how long would it take to reach 34 km (above sea level). Air resistance can be ignored.

 

Gravitational constant

 

=9.79 m/second*second

 

This is the only formulas he gave me.

 

 

Where do I even start???

 

 

Distance:

 

Distance to be traveled = 109km - 34km

Distance to be traveled = 75km

 

 

I then started making a chart going by 0.5 second increments

 

Example:

 

 

{Time} (Acceleration) [calulation]

{0.5 sec} (39.16m) [9.79/(0.5*0.5)]

{1 sec} (9.79) [9.79/(1*1)]

{1.5 sec} (4.35) [9.79/(1.5*1.5)]

{2 sec} (2.44) [9.79/(2*2)]

{2.5 sec} (1.57) [9.79/(2.5*2.5)]

{3 sec} (1.09) [9.79/(3*3)]

 

So could someone tell me is this is right

 

After 3 seconds, the rocks velocity is 58.4 M/S and traveled a distance of 156 meters

 

I got the distance by taking the time

 

{Time} (Acceleration) <Current speed> [calculation]

{0.5 sec} (39.16m) <39.16 m/s> [39.16]

{1 sec} (9.79) <48.95 m/s> [39.16+9.79]

{1.5 sec} (4.35) <53.30 m/s> [48.95+4.35]

{2 sec} (2.44) <55.75 m/s> [53.30+2.44]

{2.5 sec} (1.57) <57.32 m/s> [55.75+1.5]

{3 sec} (1.09) <58.40 m/s> [57.32+1.09]

 

I got my distance traveled by summing up my current speeds and deviding the total by 2 (2 samples per second speed is in meters per second)

 

My total speed is the total of my currentspeed

 

First thing is that correct?

 

Second thing is this is correct is there a better way of doing this?

 

If this is not correct, what is the correct calculation?

[danny8522003]

You can use Newtons Equations of motion for this as g has been given as a constant.

 

List all the things you know:

s (distance travelled) = 75,000m

a (acceleration) = 9.79ms^-2

u (initial velocity) = 0

t (time) = ?

v (final velocity) = ?

 

Here are the equations:

Equation 1 v = u + at

Equation 2 s = ½(v + u)t

Equation 3 s = ut + ½at²

Equation 4 v² = u² + 2as

 

Find one so that you input all the known values, but have one unknown and rearrange to find it.

First find v, then find t.

 

Hope that helps.

[BTICU]

Thanks, I'll give taht a shot

[danny8522003]

Ok, feel free to repost if you get stuck

[BTICU]

I'm getting a value

 

V=3791.65m/s (that sounds ok)

 

T=142,186,875 (I use equation 2)

T=0.5(3791.65+0)75,000

 

I just started algebra, can some one tell me what I'm doing wrong?

[ydoaPs]

using equation two, [math]t=\frac{2s}{v}[/math]

[BTICU]

Can anyone please veryfy these numbers

 

t=123.78s

v=1211.82m/s (please direguard the first V value)

[danny8522003]

I agree with both of those values.

 

Although, yourdadonapogo said use your value for v to find t. This is bad practice because your value for v might be incorrect, so try and use a equation to find another variable that doesnt contain one that you've had to calculate. Then there is less chance of you getting both the answers wrong .

[ydoaPs]

you can always substitute.....it will give the same answer

[Severian]

It is easier to do this using work and energy.

 

Say the distance dropped is [math]h[/math], and the downwards force is

 

[math]F=mg[/math]

 

then the work done by the force is

 

[math]W=Fh=mgh[/math]

 

 

This is the energy which is converted from potential energy to kinetic energy, so the change in kinetic energy is

 

[math]\Delta K = \frac{1}{2}mv^2 = mgh[/math]

 

So after dropping a height [math]h[/math] the downward speed is

 

[math]v=\sqrt{2gh}[/math]

 

You can then use [math]v=gt[/math] to find [math]t=v/g=\sqrt{\frac{2h}{g}}[/math]