Question

I got 48m but it's apparently wrong?
A shell is shot with an initial velocity v of 19 m/s, at an angle of θ = 57° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that air drag is negligible?

Answer 1

How did you arrive at 48m?

Answer 2

@Irishboy123
nice pic! :)

Answer 3

Hint:
Use conservation of momentum in the horizontal direction at the time of explosion.
This means that the horizontal component of velocity of one of the fragment just got faster after the explosion.
\((m_1+m_2)v_{h0}=m_1(0)+m_2(v_{h1})\)
where m1=m2=mass of each fragment.

Answer 4

So it's 50! Rounded down a couple of decimal points, but thanks! :D

Answer 5

Here are the details of the calculations:
Initially,
Velocity is v, so the horizontal and vertical components are:
\(v_h=v~cos(\theta)=10.348 m/s\), and
\(v_v=v~sin(\theta)=15.935 m/s\),

Time, t, to reach highest point
\(t=v_v~/g = 1.624 s\)
Horizontal distance travelled, S1, up to explosion
\(S_1 = v_h~t = 16.809 m\)

Since m1=m2, solving
\((m_1+m_2)v_{h0}=m_1(0)+m_2(v_{h1})\)
gives \(v_{h1}=2v_{h0}=2*10.348=20.696 m/s\).

Horizontal distance travelled, S2, from explosion to hitting ground
\(S_2 = v_{h1}*t=20.696*1.624=33.618 m\)

Total horizontal distance travelled
= 16.809+33.618
= 50.427 m.