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18 July, 17:39

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 ...?

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  1. 18 July, 19:02
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    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.
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