An archer shoots an arrow from a height 4.2 ~m above the ground with a speed 40 ~m / s and at an angle 30^ as shown in the figure. Determine the horizontal distance R covered by the arrow, when it hits the ground, (Take g=10 ~m / s ^2 )

Step-by-step solution

Given, speed of arrow, v=40 ~m / s and =30^ Horizontal range R₁ covered by the arrow is given by aligned R₁ & = v^2 2 g & = 40^2 (2 30) 10 =160 60^ & =160 3 2 =80 3 ~m aligned If t be the time taken by the arrow reaching from B to C , then from the second equation of motion, h=u t+ 1 2 g t^2 here, h=4.2 ~m , u=v 30^ 4.2=v 30^ t+ 1 2 g t^24.2=40 1 2 t+ 1 2 10 t^24.2=20 t+5 t^2 aligned & 50 t^2+200 t-42=0 & 25 t^2+100 t-21=0 aligned Solving the quadratic equation, t= 1 5 ~s Distance travelled in horizontal direction