Question

A bullet is fired on a horizontal from 1.50m. If it his a target 0.50m high that is a 100m away, how fast is the bullet traveling

Answer 1

Answer on Question #47018 – Physics – Mechanics | Kinematics | Dynamics

A bullet is fired on a horizontal from 1.50m. If it has a target 0.50m high that is a 100m away, how fast is the bullet traveling

Solution:

$\mathrm{h}_1 = 1.5\mathrm{m}$ – initial height;

$\mathrm{h}_2 = 0.5\mathrm{m}$ – final height;

$\mathrm{D} = 100\mathrm{m}$ – distance to the target;

$\mathrm{v}$ – velocity of the bullet;

$\mathrm{t}$ – time of travelling;

Equation of motion of the bullet along the X-axis:

$$\mathrm{D} = \mathrm{v} \mathrm{t}$$ $$t = \frac{D}{v}$$

Equation of motion of the bullet along the Y-axis:

$$\mathrm{h}_1 - \mathrm{h}_2 = \frac{\mathrm{g} \mathrm{t}^2}{2}$$

(1) in (2):

$$\mathrm{h}_1 - \mathrm{h}_2 = \frac{\mathrm{g} \left(\frac{\mathrm{D}}{\mathrm{v}}\right)^2}{2}$$ $$\mathrm{h}_1 - \mathrm{h}_2 = \frac{\mathrm{g} \mathrm{D}^2}{2 \mathrm{v}^2}$$ $$\mathrm{v}^2 = \frac{\mathrm{g} \mathrm{D}^2}{2 \left(\mathrm{h}_1 - \mathrm{h}_2\right)}$$ $$\mathrm{v} = \sqrt{\frac{\mathrm{g} \mathrm{D}^2}{2 \left(\mathrm{h}_1 - \mathrm{h}_2\right)}} = \sqrt{\frac{9.8 \frac{\mathrm{m}}{\mathrm{s}^2} \cdot (100 \mathrm{~m})^2}{2 (1.5 \mathrm{~m} - 0.5 \mathrm{~m})}} = 221 \frac{\mathrm{m}}{\mathrm{s}}$$

Answer: velocity of the bullet is equal to $221 \frac{\mathrm{m}}{\mathrm{s}}$.

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