Question

Two paper screens A and B are seperated by distance 100m. A bullet penetrates A and B at points P and Q respectively, where Q is 10 cm below P. If bullet is travelling horizontally at the time of hitting A, the velocity of bullet at A is

Answer 1

Answer on Question #45341, Physics, Mechanics | Kinematics | Dynamics

Two paper screens A and B are seperated by distance 100m. A bullet penetrates A and B at points P and Q respectively, where Q is 10 cm below P. If bullet is travelling horizontally at the time of hitting A, the velocity of bullet at A is

Solution.

Kinematics equations of motions:

$$O X: L = V t$$ $$O Y: h = \frac {g t ^ {2}}{2}$$

Where $t$ is the time between bullet's hits with papers. So from the second equation:

$$t = \sqrt {\frac {2 h}{g}}$$

And:

$$V = \frac {L}{t} = L \sqrt {\frac {g}{2 h}}$$

Numerically:

$$V = 1 0 0 m \cdot \sqrt {\frac {9 . 8 \frac {m}{s ^ {2}}}{2 \cdot 0 . 1 m}} = 7 0 0 \frac {m}{s}$$

Answer: $700\mathrm{m / s}$

Attachment:

Let's speak about vertical projection of bullet's speed.

Initially it was equal to 0 (point P), in the point Q it equal to:

$$V _ {y \bar {Q}} = V _ {y P} - g t = 0 - g t = - g t$$

Where $t$ is the time between bullet's hits with papers.

This vertical component is responsible for vertical displacement – value h on the picture. Let's obtain this value:

$$h _ {y} = - h = \int_ {0} ^ {t} V _ {y} (t) d t = - \int_ {0} ^ {t} g t \cdot d t = - \frac {g t ^ {2}}{2}$$

So

$$h = \frac {g t ^ {2}}{2}$$

We've just obtained the relation that initially was used to solve this problem.

In union with kinematic equation in projection on X axis its enough to obtain the answer.

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