Melody

Thanks Chris,

Do you want me to prove L'Hopital's Rule is valid for a question like this ?  

This one can be done without L'Hopital's Rule

$$\\lim \limits_{x\rightarrow 1}\;\;\frac{\sqrt{5-x}-2}{\sqrt{10-x}-3}\\\\
=lim \limits_{x\rightarrow 1}\;\;\frac{(\sqrt{5-x}-2)(\sqrt{10-x}+3)(\sqrt{5-x}+2)}{(\sqrt{10-x}-3)(\sqrt{10-x}+3)(\sqrt{5-x}+2)}\\\\
=lim \limits_{x\rightarrow 1}\;\;\frac{(5-x-4)(\sqrt{10-x}+3)}{(10-x-9)(\sqrt{5-x}+2)}\\\\
=lim \limits_{x\rightarrow 1}\;\;\frac{(1-x)(\sqrt{10-x}+3)}{(1-x)(\sqrt{5-x}+2)}\\\\
=lim \limits_{x\rightarrow 1}\;\;\frac{(\sqrt{10-x}+3)}{(\sqrt{5-x}+2)}\\\\
=\frac{(\sqrt{10-1}+3)}{(\sqrt{5-1}+2)}\\\\
=\frac{3+3}{2+2}\\\\
=\frac{6}{4}\\\\
=\frac{3}{2}\\\\$$

v=u+at

0=20+-3t

3t=20

t=20/3

t=6.6 repeater seconds

You don't  

You need to draw the picture first

you have a parabola that goes through the points (0,10), (120,10) with the vertex at (60,90)

since you have the vertex, you only need to use one of the other points.

(h,k) is the vertex so h=60, k=90

$$\\(x-h)^2=4a(y-k)\\\\
(x-60)^2=4a(y-100)\\\\
$Sub in x=0, y=10 $\\\\
(0-60)^2=4a(10-100)\\\\
3600=4a(-90)\\\\
900=a(-90)\\\\
a=-10\\\\
$So the equation is $\\\\
(x-60)^2=-40(y-100)\\\\$$

I suppose that is vertex form.  

check

V=u+10t Work out the value when u=10, t=7

V=10+10*7

you can finish it.

0^2 = 27.8^2 + 2(a)(45)

1=772.84+90a

-771.88=90a

a=-771.88/90

and you can finish it.  

Solve for x,given b=32

3^2-bx = 1

9-32*x=1

-32x=1/9

x=1/9 divided by -32

You can finish it.   

Let the co-ordinates of the start point be (0,0)

In the horizontal direction.

v=25m/s 

How long does it take the ball to travel 56.9m

$$56.9m\times \frac{1s}{25m}=2.276sec$$

so t he ball is in the air for 2.276seconds

--------------------------------------

initially in the vertical direction

t=0, v=0, a=-9.8

ongoing in the vertical direction

$$\\\ddot y=-9.8\\
\dot y=-9.8t+c_1\\
$Initial velocity = 0 so $c_1=0\\
\dot y=-9.8t\\
y=\frac{-9.8t^2}{2}+c_2\\
y=-4.9t^2+c_2\\
$When t=0, y=0 so $c_2=0\\
y=-4.9t^2\\$$

now find y when t=2.276

$${\mathtt{\,-\,}}\left({\mathtt{4.9}}{\mathtt{\,\times\,}}{{\mathtt{2.276}}}^{{\mathtt{2}}}\right) = -{\mathtt{25.382\: \!862\: \!4}}$$

So initial the ball is 25.38metres above the ground (2 dec places)

(235)^2-4(-4.9)(234)

put it into the calc on the home page as 

235^2-4*-4.9*234

you can put the brackets on if you want to.