Question

problem 4: a hunter shoots at a deer and misses. the initial speed of the bullet is $v_0 = 200 m/s$, and it is moving initially in the horizontal direction. the initial height of the bullet is $h = 1.5 m$. find the horizontal distance the bullet travelled. start with the equations for constant acceleration.

Explanation:

Step1: Analyze vertical motion

The bullet is in free - fall in the vertical direction. The initial vertical velocity $v_{0y}=0\ m/s$, the acceleration $a = g= 9.8\ m/s^{2}$, and the vertical displacement $y - y_{0}=-h=- 1.5\ m$. Using the equation $y - y_{0}=v_{0y}t+\frac{1}{2}at^{2}$, substituting the values we get $-h = 0\times t-\frac{1}{2}gt^{2}$.

Step2: Solve for time $t$

From $-h=-\frac{1}{2}gt^{2}$, we can solve for $t$. Rearranging the equation gives $t=\sqrt{\frac{2h}{g}}$. Substituting $h = 1.5\ m$ and $g = 9.8\ m/s^{2}$, we have $t=\sqrt{\frac{2\times1.5}{9.8}}\ s$.

Step3: Analyze horizontal motion

In the horizontal direction, there is no acceleration ($a_x = 0$), and the initial horizontal velocity $v_{0x}=v_0 = 200\ m/s$. The horizontal displacement $x$ is given by the equation $x = v_{0x}t$. Substitute $v_{0x}=200\ m/s$ and $t=\sqrt{\frac{2\times1.5}{9.8}}\ s$ into the equation.

Answer:

First, calculate $t=\sqrt{\frac{2\times1.5}{9.8}}\approx\sqrt{\frac{3}{9.8}}\approx0.553\ s$. Then, $x = v_{0x}t=200\times0.553 = 110.6\ m$. So the horizontal distance the bullet travelled is approximately $110.6\ m$.