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. ...?


Answer 1
Answer: 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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