# Laws of motion problem

1. Aug 12, 2014

### idontknow101

1. The problem statement, all variables and given/known data
A body of mass 50 g is dropped from a height of 20 m. A boy on the ground hits the ball vertically upwards with a bat with an average force of 200 N, so that it attains a vertical height of 45 m. The time for which the ball remains in contact with the bat is (g = 10 m/s^2)

2. Relevant equations
F= ma
v^2 = u^2 + 2gh
h = -ut +1/2 at^2

3. The attempt at a solution
v = √2gh = √2*10*20 = 20 m/s
This will become 'u' for the second part.
h = 45 m
a = F/m = 200 / 50 / 1000 = 4000 m/s^2
I substituted u, h & a in h = -ut + 1/2 at^2
I didn't get the answer which is 1/80th of a second.

I have tried using it by impulse method also.
Like, impulse = change in linear momentum = f * dt
mv - mu = f * dt
u = 20 m/s
I calculated v by v = √u^2+2gh = 10 √13
I still didn't get the answer.

I cannot understand where I am going wrong.

Last edited: Aug 12, 2014
2. Aug 12, 2014

### Orodruin

Staff Emeritus
I suggest you think a bit more about this relation. How does v relate to high the ball will go?

I also suggest you use parentheses for your square roots or use the LaTeX math mode to typeset your equations as they will be much easier to read.

3. Aug 12, 2014

### dean barry

This is how i read the sequence :
The ( 0.05 kg ) ball is dropped from rest, falling 20 metres to the first contact point ( @ 20 m/s ), the ball is then decelerated using a 200 N force, to a distance below the first contact point.
The 200 N force is continued, accelerating the ball upwards for long enough to continue on and reach 25 m above the drop point.

4. Aug 12, 2014

### Orodruin

Staff Emeritus
Yes, so how much kinetic energy does the ball have right after the contact?
How do you relate this kinetic energy to the final height the ball reaches?

(We can make the assumption that the force applies in such a short time that this does not significantly change the height of the ball - this can of course be checked later.)

5. Aug 12, 2014

### dean barry

Im thinking the decelerating force will actually be ( 200 - ( 0.05 * 10 ) ) = 199.5 N, as the mass still offers a ( gravitational ) force even while its being struck, likewise when it continues after BDC and on the way back up again.

6. Aug 12, 2014

### Orodruin

Staff Emeritus
This effect would typically be considered negligible. I would suggest trying to answer the questions I posed in #4. The first question should be answered in terms of the velocity of the ball after contact.

7. Aug 13, 2014

### dean barry

So, it decelerates from 20 m/s to 0 using a force of 200 N :
t = ( 20 * 0.05 ) / 200
t = 0.005 seconds

Then, it needs to get to a height of 45 m, so the initial velocity (u) must be :
u = sqrt ( v ² - ( 2 * a * s ) )
u = 30 m/s

And the time taken to get from 0 to 30 m/s using a force of 200 N :
t = ( 30 * 0.05 ) / 200
t = 0.0075 seconds

Total time = 0.005 + 0.0075
= 0.0080 seconds

8. Aug 13, 2014

### Orodruin

Staff Emeritus
Well, to start with: 0.005 s + 0.0075 s = 0.0125 s. However, the solution using conservation of momentum would be a simpler one than going through the actual differential equations.