How to calculatethe question?

Question

When a iarcraft is 4000m above the ground and flying upwards with a velocity of 50m/s at an angle of 30 to the horizontal.A bomb is released. Neglecting air resistance, calculate

1. the time taken by the bomb to reach the ground

2.the velocity of the bomb whwn it strikes the ground

Answer 1

Let:

u be the initial speed of the bomb,

a be the inclination of its initial path to the horizontal,

h be the initial height,

x be the horizontal distance at time t,

v be the bomb's speed on landing,

b be the inclination of its final path to the horizontal,

g be the acceleration due to gravity.

Horizontally:

x = ut cos(a) ...(1)

Vertically:

0 = h + ut sin(a) - gt^2 / 2 ...(2)

1.

Solving (2) for the positive value of t:

t = [ u sin(a) + sqrt( u^2 sin^2(a) + 2gh ) ] / g

= [ 50 sin(30) + sqrt( 50^2 sin^2(30) + 2 * 9.81 * 4000 ) ] / 9.81

= (25 + 281.256) / 9.81

t = 31.219

-> 31.2 sec.

2.

Horizontally:

v cos(b) = u cos(a) ...(3)

Vertically:

v sin(b) = - u sin(a) + gt ...(4)

Dividing (4) by (3):

tan(b) = [ gt - u sin(a) ] / [ u cos(a) ]

= [ 9.81 * 31.219 - 50 sin(30) ] / [ 50 cos(30) ]

b = 81.2 deg.

Squaring and adding (3) and (4):

v^2 = u^2 + (gt)^2 - 2ugt sin(a)

= 50^2 + (9.81 * 31.219)^2 - 2 * 50 * 9.81 * 31.219 sin(30)

v = 285 m/sec.