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.

 

[thepatientmental]

Your calculations are correct. The time taken by the bomb to reach the ground is 31 seconds and the velocity of the bomb when it strikes the ground is 263 ms-¹. Great job!