Firecracker

A firecracker explodes at a height of $\pu{100 m}$ above level ground. It breaks into a large number of very small fragments, which move in every direction; all of them have the same initial velocity of $\pu{20 m/s}$.

We assume that the fragments move without air resistance, in a uniform gravitational field with $g=\pu{9.81 m/s^2}$.

Find the volume (in $\pu{m^3}$) of the region through which the fragments move before reaching the ground.
Give your answer rounded to four decimal places.

The fragments are projected from a height \( h = 100 \, \text{m} \) with the same initial speed \( u = 20 \, \text{m/s} \) in all directions.
They travel along parabolic paths and land on the ground after some time \( t \).

We can start by finding \( t \), which is the time it takes to reach the ground again.
For this, we can use the equation of motion:
\( h = ut + 0.5gt^2 \),
which can be rearranged to:
0 = \( -0.5gt^2 + ut – h \).

Inserting the values and solving for \( t \) gives:
0 = \( -0.5 \times 9.81 \times t^2 + 20t – 100 \),
Upon solving this equation, one of the solutions will be \( t \approx 4.07 \) s,

The maximum horizontal range \( R \) a fragment can reach can be obtained as:
\( R = u \times t \).
Replacing the known values gives us:
\( R = 20 \, \text{m/s} \times 4.07 \, \text{s} \approx 81.4 \, \text{m} \).

Since the fragments spread in a circular shape, the area they cover on the ground is a circle with radius \( R \approx 81.4 \, \text{m} \), so the total surface area \( A \) will be:
\( A = \pi R^2 = \pi \times (81.4 \, \text{m})^2 \approx 20820.4 \, \text{m}^2 \).

The volume \( V \) of the space the firecracker fragments cover is shaped like a cone with height \( h = 100 \, \text{m} \) and base area \( A \approx 20820.4 \, \text{m}^2 \).

So the volume can be found as a slice (1/3) of the volume of a cylinder with height \( h \) and base area \( A \):
\( V = \frac{1}{3}Ah = \frac{1}{3} \times 20820.4 \, \text{m}^2 \times 100 \, \text{m} \approx \pu{692680 m^3} \).

So, the volume of the region through which the fragments move before reaching the ground is approximately \( \pu{692680 m^3} \). Rounded to four decimal places, this becomes \( \pu{692680.1335 m^3} \).

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