Motion of bomb?

Question

A small WWI biplane traveling at 120 km/hr in a straight and level flight 160 meters above the ground drops a small contact bomb. Find:

(a) the time it takes for the bomb to fall to the ground
(b) the horizontal distance traveled by the bomb after is it released
(c) the vertical speed acquired by the bomb as it just touches the ground
(d) the total speed of impact

Answer 1

vertical motion:
u=0, s=160m, a=10m/s^2, (a)t=? , (c)v=?
(a) using s=ut +1/2 at^2,
160=5t^2
t^2=32
t=sqrt(32)
=5.66s
v^2=u^2+2as
v^2=0+2*10*160=3200
v=sqrt(3200)
=56.57m/s
horizontzl velocity 120km/h=120000m/3600s=33.33m/s
(b)horizontal distance travelled by the bomb=hor. vel *time
= 33.33m/s*5.66s
= (10/3)*5.66
=188.7mANS.

(d) total speed of impact=sqrt[33.33^2+56.5^2]
=sqrt[1110.89+3192.25]
=65.6m/s

Answer 2

a) t=sqrt(2h/g)
t= sqrt(2 x 160 /9.81)= 5.7 s

b) S=vt
S= 120 x(1000/3600 )x 5.7=33.3 x 5.7
S= 190m
c) Vv=gt
Vv=9.81 x 5.7=55.9 m/s

d) V=sqrt(Vh^2 + Vv^2)
V= sqrt(33.3 ^2 + 55.9^2)
V=65.1 m/s