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 On a forward somersault dive, a divers height, H meters, above the water is given by h(t)= -4.9t^2+6t+3,  where T is the time in seconds after the diver leaves the board. 

 

 Determine the divers maximum height above the water. 

 

 How long does it take the diver to reach the maximum height? 

 

 For how long is the diver higher than 3 m in the water? 

 May 13, 2017
 #1
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 On a forward somersault dive, a divers height, H meters, above the water is given by h(t)= -4.9t^2+6t+3,  where T is the time in seconds after the diver leaves the board. 

 

I have already answered all this question using calculus but it occurs to me that maybe you do not know calculus yet and you were supposed to do it a different way.  Can you let me know if that is the case please.  If so I will answer again :)

 

Hi Micheala,

You need to learn to look at these and get as much information as possible just by sight.

You should recognise this as a parabola because it is a polynomial of degree 2 (the highest power is 2)

Also you can see it is concave down because the leading coefficient is negative (-4.9)

Also is you put t=0 you can see immediately that the initial height is 3 metres above the water.

 

a)  Determine the divers maximum height above the water. 

This will happen when the gradient of the tangent is 0.   And you must remember that the first derivative gives the gradient of the tangent. Find the time this happens first then sub in to find the height.

 

b)  How long does it take the diver to reach the maximum height? 

You already have that.

 

c)  For how long is the diver higher than 3 m in the water? 

Put h(t)=3 into the equation and solve it.  It is a quadratic (parabola) and you already know that one answer is t=0

find the other answer.  

 May 13, 2017

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