Answered By Majid B

components of the initial velocity:

 $v_{ix}=v_i\times\cos\theta_i=27\times\cos35=22.12$v_ix=v_i×cosθ_i=27×cos35=‍22.12 m/s

 $v_{iy}=v_i\times\sin\theta_i=27\times\sin35=15.49$v_iy=v_i×sinθ_i=27×sin35=15.49 m/s

height of travel:

 $v_{fy}^2-v_{iy}^2=2gd_y$v_ƒ y²‍−v_iy²=2gd_y     =>      $d_y=\frac{\left(v_{fy}^2-v_{iy}^2\right)}{2g}=\frac{\left(0^2-15.49^2\right)}{2\left(-9.81\right)}=12.23$d_y=(v_ƒ y²−v_iy²)2g =(0²−15.49²)2(−9.81) =12.23 m

distance of travel:  

 $g=\frac{\left(v_{fy}-v_{iy}\right)}{t}$‍g=(v_ƒ y−v_iy)t     =>     $t=\frac{\left(v_{fy}-v_{iy}\right)}{g}=\frac{\left(0-15.49\right)}{-9.81}=1.579$t=(v_ƒ y−v_iy)g =(0−15.49)−9.81 =1.579 s  =>     $T=2\times t=2\times1.579=3.158$T=2×t=2×1.579=3.158 s

 $d_x=v_{ix}\times T=22.12\times3.158=69.85$‍d_x=v_ix×T=22.12×3.158=69.85 m