Answer to Question #148057 in Physics for buddy

Question

An arrow is shot from a composite bow horizontally towards a target 105 m away. If the
arrow leaves the bow at 178 km/h, how far does it drop before it reaches the length of
the target?

Answer 1

Let's first find the flight time of the arrow from the kinematic equation:

$$x=v_0t,$$

here, $x=105\ m$ is the distance to the target, $v_0=(178\ \dfrac{km}{h})\cdot(\dfrac{1000\ m}{1\ km})\cdot(\dfrac{1\ h}{3600\ s})=49.4\ \dfrac{m}{s}$ is the initial velocity of the arrow, t is the flight time of the arrow.

Then, from this equation we can calculate $t$:

$$t=\dfrac{x}{v_0}=\dfrac{105\ m}{49.4\ \dfrac{m}{s}}=2.12\ s.$$

Finally, from the another kinematic equation we can find the arrow drop:

$$h=\dfrac{1}{2}gt^2=\dfrac{1}{2}\cdot 9.8\ \dfrac{m}{s^2}\cdot (2.12\ s)^2=22\ m.$$

Answer:

$h=22\ m.$