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4.A kicker kicks a football toward the opponent's goal line during a game. The ball begins its flight with an initial velocity of 70 ft/s when it is kicked at a height of 2 ft by the kicker. To the nearest foot, what maximum height will the ball reach? The height of the football, h,after t seconds is given by the function h(t)= 1/2 gt^2+v0t+h0. Where v0 is the initial vertical velocity, h0 is the initial height, and g=-32 ft/s^2 is the downward acceleration due to gravity.

 

my thread is locked ( i dont know how to work this site) but here my work i confused if i'm completing this right. if cant help i'll see if there some live tutoring around here. i just want to know if im setting this up right and working the problem right

 

but my answer was/

h(t)=1/2(-32)t^2+70+2

i then siimplified which i think is h(t)=-16t^2+70t+2

then i factored out a negative s16 which, hopefully looks like, y=-16(t^2- 35/8t)

my b would =b/2 which is b= -35/16 

then i square both  which is  1225/256

the add to both sides y-16*(1225/256)=-16(t^2-35/8t+1225/256)

my h,k is (35/16, 625/16)

i knew i'm supposed to end up with a parabola

 

tell me if incorrect, please.

 Nov 24, 2021
 #1
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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

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