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**Shyam** (1) a) Let the equation of the parapola is

$\displaystyle y = ax^2 + bx + c$

since it passes through (0, 0), put this in eqn, we got, c=0.

So, the eqn becomes, $\displaystyle y = ax^2 + bx$

since it passes through (10, 100),

$\displaystyle 100 = a (10)^2 + b(10)$

$\displaystyle \Rightarrow 100a + 10b = 100$

$\displaystyle \Rightarrow 10a + b = 10 ...................eqn (1)$

Since it passes through (12, 112),

$\displaystyle \Rightarrow 112 = a(12)^2 + b(12)$

$\displaystyle

\Rightarrow 144a + 12b = 112$

$\displaystyle \Rightarrow 36a + 3b = 28 .........................eqn (2)$

Now, you solve (1) and (2),to find a and b. we got $\displaystyle a = \frac {-1}{3}, \;\;b= \frac{40}{3}$

Put the value of a and b to get the equation.

equation is $\displaystyle \boxed{\;\;y = \frac{-1}{3}x^2 + \frac{40}{3}x \;\;}$

b) $\displaystyle y = \frac{-1}{3}x^2 + \frac{40}{3}x $

$\displaystyle y = \frac{-1}{3} \left(x^2 - 40x \right) $

$\displaystyle y = \frac{-1}{3} \left(x^2 - 40x + 20^2 - 20^2\right) $

$\displaystyle y = \frac{-1}{3} \left(x - 20 \right)^2 + \frac{400}{3} $

so, the maximum height reached is $\displaystyle \frac{400}{3} \;\;metres$

c) For horizontal distance, put y=0

$\displaystyle 0 = \frac{-1}{3}x^2 + \frac{40}{3}x $

$\displaystyle \Rightarrow x = 0, \;\; 40 $

Horizontal distance = 40 -0 = 40 m

Now, you try the last (d) part.