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Help me with my physics and you'll get to see a babe!!!

O

Omegachi

Diamond Member
help me with this physics problem, and i'll post up some babes 🙂 its a good deal 🙂

A projectile of maxx m = 3kg is fired with initial speed of 120 m/s at an angle of 30degrees with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 1kg and 2 kg fragment. The 2 kg fragment lands on the ground directly below the point of explosion 3.6 s after the explosion.

a. determine the velocity of the 1 kg fragment immediately after the explosion.
b. determine the energy reeleased in the explosion.



answer for part a should be :
Vx = 312 m/s
Vy = 66.6 m/s

don't know how to get those values 🙁

 


<< help me with this physics problem, and i'll post up some babes 🙂 its a good deal 🙂

A projectile of maxx m = 3kg is fired with initial speed of 120 m/s at an angle of 30degrees with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 1kg and 2 kg fragment, lands on the ground directly below the point of explosion 3.6 s after the explosion.

a. determine the velocity of the 1 kg fragment immediately after the explosion.
b. determine the energy reeleased in the explosion.



answer for part a should be :
Vx = 312 m/s
Vy = 66.6 m/s

don't know how to get those values 🙁 >>



If it moves at 312 m/s horizontally for 3.6 seconds it will land 1123.2m away. Not exactly directly below the point, unless it bounces off something, or makes some kind of weird (gravity related?) turn.
 


<< help me with this physics problem, and i'll post up some babes 🙂 its a good deal 🙂

A projectile of maxx m = 3kg is fired with initial speed of 120 m/s at an angle of 30degrees with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 1kg and 2 kg fragment. The 2 kg fragment lands on the ground directly below the point of explosion 3.6 s after the explosion.

a. determine the velocity of the 1 kg fragment immediately after the explosion.
b. determine the energy reeleased in the explosion.



answer for part a should be :
Vx = 312 m/s
Vy = 66.6 m/s

don't know how to get those values 🙁 >>


here it goes
AT maximum height the speed of the 3Kg block is 120cos30.
Now the 2kg block falls down verticalls in 3.6secs. The height of the 3Kg block when it explodes is given by 120^2sin3062/(2*9.8) = 183.7
So the speed of the 2Kg block when it explodes must be given by
-183.7 = V*3.6 -0.5*9.8*3.6^2
Solve for v = 33.3
This is the speed of the 2Kg block in Vy
Now using conservation of momentum
2*33.3 = 1*Vy -> Vy = 66.6

Now about the horizontal velocity
Use conservation of momentum
Intial momentum is 3*120*cos30
FInal momentum 1*Vx
Vx = 3*120*cos30 = 311.76m/s
Energy released = 0.5*1*(312^2+66.6^2)+0.5*2*(33.3^2) - 0.5*3*(120cos30)^2=calculate it urself buddy
enjoy!!!
 
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