# Projectile Motion (potential & kinetic energy)

1. Nov 11, 2012

### Arty7

The question is in the attachment.

a) for the height i got 1.25m which you get by using displacement and acceleration. (t=20/40=0.5sec)

b) a=v/t, 10*0.5=5m/s

c) At initial point:

mgh= 0.040*10*1.25=0.5J
1/2mv^2=1/2*0.040*40^2=32J

Hits Ground:
mgh=0.040*10*0=0J
1/2mv^2=1/2*0.040*(20 or 40)^2 ?????? The answer is 32.5J im not sure what happens here and how to get that. Do you add the energys from initial point together???

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Last edited: Nov 11, 2012
2. Nov 11, 2012

### lewando

Since there are no losses due to wind resistance the initial energy equals the final energy. The velocity you use right before it hits ground is the composite velocity.

3. Nov 11, 2012

### Arty7

sorry i dont understand that very well.

4. Nov 11, 2012

### tiny-tim

Hi Arty7!
Use components.

5. Nov 11, 2012

### Arty7

so.. mgh+1/2mv^2 0.5 +32 =32.5J

6. Nov 11, 2012

### lep11

As for b use vy=g*t

7. Nov 11, 2012

### Abhinav R

a) Use the formula : h=(1/2)gt²
h=1$/$2 $\times$ 10$\times$(0.5)²
Here I calculated t from the usual t = d$\div$speed
t=0.5 s
Coming back to h,We get h as 1.25 m

b) v$_{x}$ = u = 40 m/s
v$_{y}$ = u$_{y}$+gt=0+10$\times$0.5
= 5 m/s

c) KE$_{initial}$ = (1$\div$2)$\times$m$\times$v²
m=40 g and v= 40 m/s
= 0.5 J
KE$_{final}$ = (1$\div$2)$\times$m$\times$v$_{final}$²
Here you need to find the final velocity v$_{final}$ ; Find by v = u + gt
u=40 m/s and t=0.5 s
So v = 45 m/s
∴ KE$_{final}$ = 0.9 J if I'm right
Now KE$_{final}$ - KE$_{initial}$ = 0.4 J

Similarly Potential Energy $\rightarrow$
PE$_{initial}$ = mgh
$\rightarrow$40 × 10$^{-3}$ × 10 × 1.25
= 0.5 J
PE$_{final}$ = mgh
Here h is the equation of the trajectory,
We get h as 6.25 × 10$^{-3}$
Now the potential energy is 2.5 × 10$^{-4}$ J
I guess PE(final) - PE(inital) = 2 × 10$^{-4}$ J
Now PE + KE = Answer
PE + KE = ?

Last edited: Nov 11, 2012
8. Nov 11, 2012

### lewando

I think you meant:
v=$\sqrt{v^{2}_{x} + v^{2}_{y}}$

Last edited: Nov 11, 2012