# Thread: Anyone good with Physics?....im sure!

1. ## Anyone good with Physics?....im sure!

force acting on a mass of 50kg when it will will accel' it from 2m/s to 9m/s in 10s?.......I got 9-2 = 7, f=ma, (50x7) = 350n Is this right?

the one im stuck on though, is a ball of mass 5kg launched upwards with 800n force gets to height 'H'. What is 'H' and what is the time from launch until ball returns to ground. (when g = 9.8m's^2)

I know this doesn't quite belong in maths forum but you guys have ben a big help in the past, any input would be much appreciated. THANKS!

2. Originally Posted by bobchiba
the one im stuck on though, is a ball of mass 5kg launched upwards with 800n force gets to height 'H'. What is 'H' and what is the time from launch until ball returns to ground. (when g = 9.8m's^2)
It depends on how long the force acts for.

800n will accelarate a 5kg mass upwards at a=800/5 -g m/s^2

Now if we were talking about an impulse of 800 newton seconds we could do
something.

RonL

3. Whooops....sorry, your right, it acts for 20 seconds.

4. Originally Posted by bobchiba
the one im stuck on though, is a ball of mass 5kg launched upwards with 800n force gets to height 'H'. What is 'H' and what is the time from launch until ball returns to ground. (when g = 9.8m's^2)
Originally Posted by bobchiba
Whooops....sorry, your right, it acts for 20 seconds.
First a note: The unit "Newton" is named after someone, so it should properly be abbreviated with "N," not "n."

F = ma
800 N = (5 kg)*a

a = 800/5 m/s^2 = 160 m/s^2

The net acceleration is then
160 m/s^2 - 9.8 m/s^2 = 150.2 m/s^2 (upward)<-- Since the object's weight still acts on it.

So, what is the equation of motion of an object under a constant acceleration?
y = y0 + v0*t + (1/2)a*t^2
I'll set the origin at the point where the ball lifted off and +y upward. So y0 = 0 m and v0 = 0 m/s (presumably it was launched from rest, the problem never actually stated this.) So at the end of 20 s the height H will be:

H = (1/2)a*t^2
H = (1/2)*150.2*(20)^2 m = 30040 m.

To find out the time from launch to when it hits the ground we need two pieces of information: how long it takes for the ball to reach the ground from the height H, and what the velocity is at the end of the 800 N acceleration period. We'll do this part first.

While the ball is accelerated upward we have:
v = v0 + at <-- The ball starts from rest, so v0 = 0 m/s.
v = 150.2*20 m/s = 3004 m/s
(Note: since the sign on this velocity is + the velocity is upward.)

Now I'm going to reset the problem such that the time starts at the end of the 800 N acceleration phase. The ball starts at a height of 30040 m with an (upward) initial velocity of 3004 m/s and an acceleration only due to gravity of -9.8 m/s^2 (negative because it's acting downward.) The equation of motion is:
y = 30040 + 3004t - (1/2)*9.8*t^2

We wish to know when y = 0 m again. (Then add 20 s to this time and we get our final answer.) So:
-4.9t^2 + 3004t + 30040 = 0
Using the quadratic equation I get:
t = 622.903 s or t = -9.842 s
The negative time is clearly a ridiculous answer, so it takes 622.903 s to get back to the ground from this height.

Thus the total time of flight is
622.903 s + 20 s = 642.903 s

-Dan