Kinematics: Calculate Time & Velocity of Bomb from 4000m Altitude

[bADc]

i have a question here about kinematics.

1. When a aircraft is 4000m above the ground and flying upwards with a velocity of 50 ms-¹ at an angle of 30º to the horizontal. A bomb is released.Neglecting air resistance, calculate

(a)the time taken by the bomb to reach the ground,
(b)the velocity of the bomb when it strikes the ground.(g=10 ms-²)

(a)h=4000m
vertical component of velocity=25.5
horizontal component of velocity=43.3

s=ut+1/2at²
4000=0+5t²
t=28.5

v=u+at
25=0+10t
t=2.5

t=28.5+2.5
t=31
is it correct?

(b)vertical component of velocity=25.5

v=u+at
=0+10(28.5)
=285 ms-¹

vertical component of velocity=285 ms-¹-25.5 ms-¹=259.5 ms-¹

velocity of bomb, v=(259.5²+43.3²)½
=263 ms-¹

please someone check the answer 4 me.TQ!

 

[theCandyman]

Edit: Sorry I misinterpeted your work. Everything looks in order.