Answer:
Range of arrow = 225.09 meter
Final horizontal velocity = 34.47 m/s
Explanation:
We have equation of motion [tex]s=ut+\frac{1}{2} at^2[/tex], where u is the initial velocity, t is the time taken, a is the acceleration and s is the displacement.
Considering the vertical motion of arrow ( up direction as positive)
We have u = 45 sin40 = 28.93 m/s, s = -20 m, a = acceleration due to gravity = [tex]-9.8m/s^2[/tex].
[tex]-20=28.93*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-28.93t-20=0[/tex]
t = 6.53 seconds or t = -0.63 seconds
So time = 6.53 seconds.
Considering the horizontal motion of arrow
u = 45 cos 40 = 34.47 m/s, t = 6.53 s, a = [tex]0m/s^2[/tex]
[tex]s=34.47*6.53+\frac{1}{2} *0*6.53^2\\ \\ s=225.09m[/tex]
So range of arrow = 225.09 meter
Horizontal velocity will not change , final horizontal velocity = 34.47 m/s.
