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

Options

A. `(185)/(sqrt3)m`
B. `84sqrt3m`
C. `68sqrt3m`
D. `(95)/(sqrt3)m`

Answer

B

Explanation

Given, speed of arrow,
`v=40m//sandtheta=30^(@)`

Horizontal range `R_(1)` covered by the arrow is given by
`R_(1)=(v^(2)sin2theta)/(g)`
` =(40^(2)sin(2xx30))/(10)=160xxsin60^(@)`
`=160(sqrt3)/(2)=80sqrt3m`
If t be the time taken by the arrow reaching from B to C, then from the second equation of motion,
`impliesh=ut+(1)/(2)g""t^(2)`
here, h=4.2m, `u=vsin30^(@)`
`4.2=vsin30^(@)t+(1)/(2)g""t^(2)`
`4.2=40xx(1)/(20)t+(1)/(2)xx10t^(2)`
`4.2=20t+5t^(2)`
`50t^(2)+200t-42=0`
`25t^(2)+100t-21=0`
Solving the quadratic equation, `t=(1)/(5)s`
Distance travelled in horizontal direction is given by
`R_(2)=vcos30^(@)xxt`
`=40(sqrt3)/(2)xx(1)/(5)=4sqrt3m`
Hence, the horizontal distance covered by the arrow `R=R_(1)+R_(2)`
`=80sqrt3+4sqrt3)=84sqrt3m`