Question

An archer fires an arrow towards the wall of a castle. The arrow is fired at 45° from horizontal at a velocity of 22m/s. If the wall is 40m away and stands 7m tall, calculate whether the arrow will drop inside the castle walls.


a) Calculate the maximum height of the arrow

b) Calculate the maximum range of the arrow

c) Calculate the vertical height of the arrow as it reaches the wall

d) Describe the journey of the arrow and whether or not it was successful

Answer 1

$\alpha =45^{\circ},\; v_0=22m/s, \; l_0=40m,\; h_0=7m$

Answers:

a)$h=12.347m$

b)$l=49.387m$

c)$H=7.603m$

d)The arrow will drop inside the castle walls

Explanation:

$x=v_{0x}t, \; v_{0x}=v_0 \cos \alpha$

$y=v_{0y}t-\frac{gt^2}{2}, \; v_{0y}=v_0 \sin \alpha$

$v_x=v_{0x}, \; v_y=v_{0y}-gt$

a) Let $t_1$ be time point, when the arrow reaches the maximum height.

At this point $v_y=0, \quad v_0 \sin \alpha -gt_1=0, \quad t_1=\frac{v_0 \sin \alpha}{g}$

$h=v_{0y}t-\frac{gt^2}{2}=v_0 \sin \alpha \frac{v_0 sin \alpha}{g} -\frac{g}{2}(\frac{v_0 \sin \alpha}{g})^2=\frac{v_0^2{\sin ^2\alpha}}{2g}$

$h= \frac{v_0^2{\sin ^2\alpha}}{2g}= \frac{22^2 * \frac{1}{2} }{2*9.8} \approx 12.347m$

b) Let $t_2$ be time point, when the arrow reaches the maximum range.

$l=v_{0x} t_2=v_0 \cos \alpha \; 2t_1=v_0 \cos \alpha \; \frac{2v_0\sin\alpha}{g}=\frac{v_0^2 \sin 2\alpha}{g}$

$l =\frac{v_0^2 \sin 2\alpha}{g}=\frac{22^2 *1}{9.8} \approx49.387m$

c) Let $t_3$ be time point, when the arrow reaches the wall.

$l_0=v_{0x}t_3$

$H=v_{0y}t_3-\frac{gt_3^2}{2}= v_0 \sin \alpha \frac{l_0}{v_0 \cos \alpha} -\frac{gl_0^2}{2v_0^2 \cos ^2 \alpha}= l_0 \tan \alpha-\frac{gl_0^2}{2v_0^2 \cos ^2 \alpha}=40*1-\frac{9.8*40^2}{2*22^2 1/2}=7.603m$

d) $H>h_0 \;\Rightarrow$ the arrow will be under the wall