Answered By Majid B

The distance ball travels horizontally:

v_ix = v_i .  $\cos\theta_i=20\times\cos30=17.32$cosθ_i=20×cos30=17.32  m/s

d_x = v_ix . t = 17.32$\times3=51.96$×‍3=51.96 m 

R = 51.96 m

The height from which ball was thrown:

v_iy = v_i . $\sin\theta_i=20\times\sin30=10$sinθ_i=20×sin30=10   m/s

d_y = (1/2).a_y.t²+v_iy.t = (1/2) (-9.81) (3²)+(10)(3)=-14.15 m

h₀ = 14.15 m

The maximum height of the ball:

v_fy² - v_iy² = (2) a_y . d_y

0² - 10² = (2) (-9.81) d_y

d_y =  $\frac{-100}{-19.62}=5.10$−100−19.62 =5.10 m

h_max = 5.10 + 14.15 = 19.25 m