Question

A bullet is fired at a 10° angle from a rifle having a muzzle velocity of 150m/sec.
How many seconds is the bullet un the air?
How far does the bullet travel horizontally before striking the ground?
How far does the bullet Rose before beginning to fall?

Answer 1

Answer on Question #71521, Physics / Mechanics | Relativity

Question. A bullet is fired at a $10{}^{\circ}$ angle from a rifle having a muzzle velocity of $150 \, m/sec$. How many seconds is the bullet in the air? How far does the bullet travel horizontally before striking the ground? How far does the bullet rose before beginning to fall?

Given. $\alpha = 10{}^{\circ}; v_0 = 150 \, m/s$.

Find. $t, S, h_{max} - ?$.

Solution.

The initial horizontal velocity is $v_{0x} = v_0 \cos \alpha$ and the initial vertical velocity is $v_{0y} = v_0 \sin \alpha$. So

$$y = v_{0y} t - \frac{g t^2}{2} \quad \text{and} \quad v_y = v_{0y} - g t = v_0 \sin \alpha - g t$$

If $y = h_{max}$ then $v_y = 0$. Hence

$$t_1 = \frac{v_{0y}}{g} = \frac{v_0 \sin \alpha}{g}, \quad t_1 = t_2 \rightarrow t = t_1 + t_2 = \frac{2 v_0 \sin \alpha}{g} = \frac{2 \cdot 150 \cdot \sin 10}{9.8} = 5.3 \, \text{s}.$$

$t_1$ is the time during which the bullet was moving upwards; $t_2$ is the time during which the bullet was moving down.

$$S = v_{0x} t = v_0 \cos \alpha \frac{2 v_0 \sin \alpha}{g} = \frac{v_0^2 \sin 2\alpha}{g} = \frac{150^2 \cdot \sin 20{}^{\circ}}{9.8} = 785 \, \text{m}.$$ $$y = h_{max} = v_{0y} t_1 - \frac{g t_1^2}{2} = v_0 \sin \alpha \cdot \frac{v_0 \sin \alpha}{g} - \frac{g}{2} \left(\frac{v_0 \sin \alpha}{g}\right)^2 = \frac{v_0^2 \sin^2 \alpha}{2g} = \frac{150^2 \cdot \sin^2 10{}^{\circ}}{2 \cdot 9.8} = 34.6 \, \text{m}.$$

Answer. $t = 5.3 \, \text{s}; S = 785 \, \text{m}; h_{max} = 34.6 \, \text{m}$.

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