Projectile Motion (potential & kinetic energy)

[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 I am not sure what happens here and how to get that. Do you add the energys from initial point together?

 

[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.

 

[Arty7]

sorry i don't understand that very well.

 

[tiny-tim]

Hi Arty7!

1/2mv^2=1/2*0.040*(20 or 40)^2 ?

Use components.

 

[Arty7]

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

 

[lep11]

As for b use v_y=g*t

 

[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 = ?

 

[lewando]

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