Melody

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UsernameMelody
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 #1
avatar+115358 
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Thank you for finally putting some effort into your presentation.

 

A mistake that you have made is thinking that    V0 is 70

The initial velocity has both a vertical velocity element and a horizontal velocity element.

The vertical velocity element is   V0 is  70sin(theta)  where theta is the initial angle of the trajectory (the angle of elevation)

 

So 

\(h(t)= 1/2 gt^2+v_0t+h_0\\ h(t)=-16t^2+(70sin\theta)t+2\)

 

You are right, this is a concave down parabola and you can find the maximum height by finding h at the centre point.

 

Centre = -b/2a   from the quadratic formula

\(t=\frac{-70sin\theta}{-32}=\frac{35sin\theta}{16}\\~\\ h(\frac{35sin\theta}{16})=-16(\frac{35^2sin^2\theta}{16*16})+70sin\theta*\frac{35sin\theta}{16}+2\\ h(\frac{35sin\theta}{16})=-(\frac{35^2sin^2\theta}{16})+\frac{70*35sin^2\theta}{16}+2\\ h(\frac{35sin\theta}{16})=(\frac{35^2sin^2\theta}{16})+2\\ h(\frac{35sin\theta}{16})=(\frac{35sin\theta}{4})^2+2\\ \)

This is the maximum height that will be reached.  You cannot give it to the nearest foot because you have not been given the initial angle. 

And you have not been given the initial vertical velocity.  If the question is worded incorrectly and 70 was supposed to be the initial vertical velocity then the logic behind what you have done looks fine.

 

 

Please note:  All answers provided need to be checked for logic as well as careless errors.

 

 

LaTex:

t=\frac{-70sin\theta}{-32}=\frac{35sin\theta}{16}\\~\\
h(\frac{35sin\theta}{16})=-16(\frac{35^2sin^2\theta}{16*16})+70sin\theta*\frac{35sin\theta}{16}+2\\
h(\frac{35sin\theta}{16})=-(\frac{35^2sin^2\theta}{16})+\frac{70*35sin^2\theta}{16}+2\\
h(\frac{35sin\theta}{16})=(\frac{35^2sin^2\theta}{16})+2\\
h(\frac{35sin\theta}{16})=(\frac{35sin\theta}{4})^2+2\\

Nov 24, 2021