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 $y=-h = - 1.5\ m$ (taking downwards as negative). Using the equation $y = v_{0y}t+\frac{1}{2}at^{2}$, substituting the values we get $-1.5=0\times t+\frac{1}{2}\times(- 9.8)t^{2}$.

Step2: Solve for time $t$

From $-1.5 = - 4.9t^{2}$, we can solve for $t$. Rearranging the equation gives $t^{2}=\frac{1.5}{4.9}$, so $t=\sqrt{\frac{1.5}{4.9}}\ s$.

Step3: Analyze horizontal - motion

In the horizontal direction, there is no acceleration ($a_x = 0$), and the initial horizontal velocity $v_{0x}=200\ m/s$. The horizontal distance $x$ is given by the equation $x = v_{0x}t$. Substituting $v_{0x}=200\ m/s$ and $t=\sqrt{\frac{1.5}{4.9}}\ s$ into the equation, we have $x = 200\times\sqrt{\frac{1.5}{4.9}}$.
Calculating $x = 200\times\sqrt{\frac{1.5}{4.9}}\approx200\times0.553\approx110.6\ m$.

Answer:

$110.6\ m$