A medieval prince trapped in a castle wraps a message around a rock and throws it from the top of the castle with an initial velocity of 12m/s[42 degrees of above the horizontal]. The rock lands just on the far side of the castle's moat, at a level 9.5m below the initial level. Determine the rock's time of flight ...?

Vertically we can say

Vertical acceleration = - g

Vertical velocity = u sinѲ - gt [Ѳ = 42; u = 12 m/s]

Vertical displacement = u sinѲt - (1/2) g t^2 + 9.5

When the rock hits the ground, its vertical displacement will be zero. So we can say ...

u sinѲt - (1/2) g t^2 + 9.5 = 0

I'll rearrange this ...

- (1/2) g t^2 + u sinѲt + 9.5 = 0

Can you see that we now have a quadratic in t?

Using the well known formula

t = [ - u sinѲ ± √ (u^2 sin^2Ѳ - 4 (( - (1/2) g * 9.5)) ] / ( - g)

t = [ - u sinѲ ± √ (u^2 sin^2Ѳ + 19g) ] / (-g)

t = [ - 8.03 ± √ (250.86) ] / (-9.81)

t = [ - 8.03 ± 15.84] / (-9.81)

t = ( - 8.03 + 15.84) / (-9.81) or t = (-8.03 - 15.84) / (-9.81)

t = - 0.8 or t = 2.43

Well, a negative time has no meaning so

t = 2.43 seconds.