initial velocity v = 86.6 m / s angle ,theta = 63.7 degrees (A). the horizontal component of the shell's initial velocity = v cos 63.7 = 38.36 m / s vertical component of shells' initial velocity = v sin 63.7 = 77.63 m / s (B). time taken to reach maximu height t = v sin (theta ) / g = 7.922 s (C). Maximum height H = ( v sin(theta) )^ 2/ 2g = 307.5 m (D). Required distance = v^ 2 *sin2(theta) / g = 607.9 m (E). Horizontal component of accleraton = 0 vertical component of accleration = g = 9.8 m / s^ 2 (F). horizontal component of velocity at required point = v cos (theta ) = 38.36 m / s vertical compoent of velocity at required point = 0
