Physics time/horizontal distance?

Question

A seed shoots out from the pod with the speed of 3.2 m/s but with a direction of motion 30° below the horizontal. The seed pod is initially 0.9 m above the ground.

(a) How long does it take for the seed to land? (Neglect air resistance.)

(b) What horizontal distance does it cover during its flight?

Answer 1

Let:

u be the initial speed

t be its angle below the horizontal,

y be the initial height of the seed above ground,

x be the horizontal distance travelled,

g be the acceleration due to gravity.

Horizontallty:

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

Vertically:

0 = y - ut sin(a) - gt^2 / 2

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

From (2), the positive root is:

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

= [ - 3.2 sin(30) + sqrt(3.2^2 sin^2(30) + 2 * 9.81 * 0.9) ] / 9.81

= 0.295 s.

Substituting this value of t in (1):

x = 0.818 m.