Question

what is the range of a bullet fired horizontally at a height of 1.5 m with an initial velocity of 120 m/s? (1 point)
≈66.4 m
-9.8 m/s²
≈0.55 s
there is not enough information to find the range of the bullet.

Explanation:

Step1: Find the time of flight

Use the vertical motion equation \(y = v_{0y}t+\frac{1}{2}gt^{2}\). Since \(v_{0y} = 0\) (fired horizontally), \(y=- 1.5\ m\) (downward direction) and \(g=-9.8\ m/s^{2}\).

$$y=\frac{1}{2}gt^{2}$$

$$t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2\times1.5}{9.8}}$$

$$t=\sqrt{\frac{3}{9.8}}\approx0.55\ s$$

Step2: Find the horizontal range

Use the horizontal motion equation \(x = v_{0x}t\). Here \(v_{0x}=120\ m/s\) and \(t = 0.55\ s\)

$$x=120\times0.55 = 66\ m\approx66.4\ m$$

Answer:

\(\approx66.4\ m\)