Question

The horizontal range of a bullet fired with angle of projection 45 degree to the horizontal is 360 metres if it is fired from a lorry movie in the direction of bullet with the uniform velocity 18 km per hour and with same elevation, what is the new range horizontal distance travelled by the bullet ?

Answer 1

Answer on Question #76298, Physics Mechanics Relativity

The horizontal range of a bullet fired with angle of projection 45 degree to the horizontal is 360 metres if it is fired from a lorry movie in the direction of bullet with the uniform velocity 18 km per hour and with same elevation, what is the new range horizontal distance travelled by the bullet?

Solution.

The projection on the axis X: $v_{0x} = v_0 \cdot \cos \alpha$

The projection on the axis Y: $v_{0y} = v_0 \cdot \sin \alpha$

Flight time: $t = \frac{2 \cdot v_0 \cdot \sin \alpha}{g}$

Range horizontal distance travelled by the bullet: $L = \frac{v_0^2 \cdot \sin 2\alpha}{g}$

$$v_0 = \sqrt{\frac{L \cdot g}{\sin 2\alpha}} = \sqrt{\frac{360 \cdot 10}{\sin 90{}^\circ}} = 60 \frac{m}{s}$$

If it is fired from a lorry movie in the direction of bullet with the uniform velocity 18 km per hour and with same elevation:

18 km per hour = 5 metres per second

$$v_0' = v_0 + 5 = 60 + 5 = 65 \frac{m}{s}$$ $$L = \frac{v_0'^2 \cdot \sin 2\alpha}{g} = \frac{(65)^2 \cdot \sin 90{}^\circ}{10} = \frac{4225 \cdot 1}{10} = 422.5 \, m$$

Answer: $L = 422.5 \, m$

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