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## Homework Statement

A 12 kg shell is launched at an angle of 64 degrees above the horizontal with an initial speed of 120 m/s. When it is at its highest point, the shell exploded into two fragments, one three times the mass of the other. The two fragments reach the ground at the same time. You can ignore air resistance. The heavier fragment lands back at the original launching point.

How far from the launch does the lighter fragment land?

How much energy is released in the explosion?

## Homework Equations

p=mv

k=.5mv^2

x=vt

vf=vi+at

## The Attempt at a Solution

V(horizontal)=52.604m/s

V(Vertical)=107.885m/s

KE(Vertical)= .5mv(vertical)= 69796.576

KE(Horizontal)= .5mv(horizontal)= 16603.424

KE(start) = PE(max height)

69796.576 = mgh

h = 593.5

vf=vi + at

t=11.006

x = vt

x = 578.941

If b = mass of small piece

b + 3b = 12

b = 3

small mass = 3

big mass = 9

KE(Horizontal) = .5mv^2

16603.424=.5(3)v^2

v=105.21

x = vt

x = 1157.93

total x = 1157.93 + 578.941 = 1736.87m

annnd its wrong