Question

Physics Kinematics question

Answer 1

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)