Question

An archer shoots an arrow with a velocity of 30m/s at an angle of 20 degrees with respect to the horizontal. An assistant standing on the level ground 30m downrange from the launch point throws an apple straight up with the minimum initial speed necessary to meet the path of the arrow. What is the initial speed of the apple and at what time after the arrow is launched should the apple be thrown so that the arrow hits the apple?

Answer 1

we need to set up parametric equations for both the arrow and the apple
apple:
\[x(t_a) = 30\]
\[y(t_a) = -4.9 t_a^2 + v t_a\]

arrow:
\[x(t) = 30\cos (20) t\]
\[y(t) = -4.9t^2 + 30 \sin(20) t\]
to find point of intersection, set components equal:
by setting x(t) = x(t), you can find "t" or time it takes for arrow to go 30 m
\[30 \cos(20) t = 30 \rightarrow t = \frac{1}{\cos (20)}\]
Now we can plug this "t" value in to find the height of arrow when it is directly over the apple
\[y = -4.9 (\frac{1}{\cos^2 (20)}) + 30 \sin(20) (\frac{1}{\cos(20)})\]
\[y = -4.9 \sec^2 (20) + 30 \tan(20)\]
The min velocity would be just enough so that the apple reaches this height at its apex
apex of parabola is midpoint between intercepts
\[\rightarrow -4.9 t_a ^2 + vt_a = 0\]
\[t (-4.9t +v) = 0 \rightarrow t = 0, t = \frac{v}{4.9}\]
midpoint = v/9.8
so when t_a = v/9.8 the height should equal height of arrow
\[-4.9 (\frac{v}{9.8})^2 + v(\frac{v}{9.8}) = -4.9 \sec^2 (20) + 30 \tan(20)\]
solve for "v"
\[v = 2 \sqrt{4.9} \sqrt{-4.9 \sec^2(20) +30\tan(20)}\]
the time apple should be thrown is:
\[t - t_a = \frac{1}{\cos(20)} - \frac{v}{9.8}\]