Question

On level ground a shell is fired with an initial velocity of 46.0 m/s at 66.0 ∘ above the horizontal and feels no appreciable air resistance.

Answer 1

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