An interesting problem, and thanks to the precise heading you put for the question.
We will assume zero air resistance. We further assume that the angle with vertical is t=53.13 degrees, corresponding to sin(t)=0.8, and therefore cos(t)=0.6.
Given: angle with vertical, t = 53.13 degrees sin(t)=0.8; cos(t)=0.6; air-borne time, T = 20 seconds initial height, y0 = 800 m
Assume g = -9.81 m/s^2
initial velocity, v m/s (to be determined)
Solution:
(i) Determine initial velocity, v. initial vertical velocity, vy = vsin(t)=0.8v Using kinematics equation, S(T)=800+(vy)T+(1/2)aT^2 ....(1) Where S is height measured from ground.
substitute values in (1): S(20)=800+(0.8v)T+(-9.81)T^2 => v=((1/2)9.81(20^2)-800)/(0.8(20))=72.625 m/s for T=20 s
(ii) maximum height attained by the bomb Differentiate (1) with respect to T, and equate to zero to find maximum dS/dt=(vy)+aT=0 => Tmax=-(vy)/a = -0.8*72.625/(-9.81)= 5.9225 s
Maximum height, Smax =S(5.9225) =800+(0.8*122.625)*(5.9225)+(1/2)(-9.81)(5.9225^2) = 972.0494 m
(iii) Horizontal distance travelled by the bomb while air-borne Horizontal velocity = vx = vcos(t) = 0.6v = 43.575 m/s Horizontal distace travelled, Sx = (vx)T = 43.575*20 = 871.5 m
(iv) Velocity of the bomb when it strikes ground vertical velocity with respect to time V(T) =vy+aT...................(2) Substitute values, vy=58.1 m/s, a=-9.81 m/s^2 V(T) = 58.130 + (-9.81)T => V(20)=58.130-(9.81)(20) = -138.1 m/s (vertical velocity at strike)
vx = 43.575 m/s (horizontal at strike) resultant velocity = sqrt(43.575^2+(-138.1)^2) = 144.812 m/s (magnitude) in direction theta = atan(43.575,138.1) = 17.5 degrees with the vertical, downward and forward. (direction)
