Projectile explodes at peak where will it fall?

[jaymode]

There is a projectile that explodes into equal mass fragments at its highest point. One falls vertically with initial speed. I need to find where the other will strike the ground and the energy released during the explosion.

I am given the mass(19.6kg), angle (56 degress), initial velocity(79.0m/s), and g (9.8m/s^2).

I know i need to use center of mass somehow but I really do not know how to go about solving this and could use some help.

 

[stunner5000pt]

is this a projectile which was shot at an angle of 56 degrees with velocity 79m/s and then exploded at the highest point??

In that case you can use conservation of momentum for this one fragment that bursted into two with one going down and the other going off at an angle.

Draw a diagram of this and figure it out
The momentum is conserved only for the initial moments so eveything you calculate is the instantaneous (at the highest point) velocity. Thereafter you can calculate parts of the second piece that went at the angle upward

 

[thepatientmental]

To determine where the other fragment will strike the ground and the energy released during the explosion, we can use the principle of conservation of momentum and energy.

First, let's label the two fragments as A and B. Fragment A falls vertically with an initial speed of 79.0m/s. Using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is acceleration (in this case, acceleration due to gravity which is -9.8m/s^2), and t is time, we can find the time it takes for fragment A to reach the ground.

v = u + at
0 = 79.0 - 9.8t
t = 8.06 seconds

Therefore, fragment A will take 8.06 seconds to reach the ground.

Now, let's look at fragment B. Since the explosion happens at the peak of the projectile's trajectory, we can assume that fragment B has the same initial velocity and angle as the projectile. Using the formula v = u + at, we can find the horizontal and vertical components of fragment B's velocity.

Horizontal component:
v_x = u_x = u * cos(56) = 79.0 * cos(56) = 42.50m/s

Vertical component:
v_y = u_y + at
v_y = u * sin(56) - 9.8 * 8.06
v_y = 79.0 * sin(56) - 78.87
v_y = 50.56m/s

Now, we can use the formula x = ut + 1/2at^2 to find the horizontal distance traveled by fragment B before it reaches the ground.

x = ut + 1/2at^2
x = 42.50 * 8.06 + 1/2 * 0 * (8.06)^2
x = 342.65m

Therefore, fragment B will strike the ground at a horizontal distance of 342.65m from the point of explosion.

To calculate the energy released during the explosion, we can use the formula KE = 1/2mv^2, where KE is kinetic energy, m is mass, and v is velocity.

For fragment A:
KE_A = 1/2 * 19.6 * (79.0)^2
KE_A = 61,726.80 Joules