Question

A 3-kg projectile is launched at an angle of 45o above the horizontal. The projectile explodes at the peak of its flight into two pieces. A 2-kg piece falls directly down and lands exactly 50 m from the launch point. Determine the horizontal distance from the launch point where the 1-kg piece lands.

Answer 1

1517.4 m

Step-by-step Step-by-step explanation:

Since the projectile broke up at the peak of its flight, it already traveled half its initial range so we can find its initial launch velocity
`v_0` from the equation

`(1)/(2)R= (1)/(2) \left((v_0^2)/(g)\sin 2\theta_0 \right)`

where
`\theta_0 = 45°` and
`(1)/(2)R = 50\:\text{m}` so we will get
`v_0=31.3\:\text{m/s}`. Next, we can use the equation

`v_y = v_0y - gt = v_0 \sin 45 - gt`

and since
`v_y=0` at its peak, we get t = 22.1 s. Let's set this aside for a moment and we'll use it later.

At the top of its peak, we can use the conservation law of linear momentum. Let M be the mass if of the original projectile,
`m_1` be the mass of the larger fragment (2 kg) and
`m_2` be the mass of the smaller fragment (1 kg). We can write the conservation law as

`Mv_0x = m_1V_1 + m_2V_2`

where
`V_1\:\text{and}\:V_2` are the velocities of the fragments immediately after the break up. But we also know that
`V_1=0` so the velocity of
`m_2` can be calculated from the conservation law as

`Mv_0 \cos 45° = m_2V_2`

or

`V_2 = (M)/(m_2)v_0 \cos 45° = 66.4\:\text{m/s}`

Now we can calculate the horizontal distance the smaller fragment traveled after the break up. Recall that the amount of time for it to go up is also the amount of time to get down so the horizontal distance x is

`x = V_2 t = (66.4\:\text{m/s})(22.1\:\text{s})= 1467.4\:\text{m}`

Therefore, the total distance traveled from the launch point is

`D = 50\:\text{m} + 1467.4\:\text{m}=1517.4\:\text{m}`