Question

The trajectory of a shot put of an athlete is modelled by a quadratic graph. The shot was thrown off the hand of the athlete at a height of 2 metres. The max. point of the trajectory was at (3, 2.5).. ?

(a) Express the equation of the trajectory in the form y=a(x-h)^2 + k, where a, h, and k are constants. (b) The trajectory of another attempt of the shot put by the athlete had a y-intercept of 2, a max. point at x= 4 and a horizontal range of 10 m. Find the equation for this trajectory. ?

Answer 1

For (a)`=-1/18 (x-3)^2+2.5`

For (b)`=-1/10 (x-4)^2+3.6`

Explanation:

Given, height =2 metres. Since it is the initial height when no distance is covered. So x=0 y=2
An y intercept.
Max at `(3,2.5)`.

Let the equation be,
`y=a (x-h)^2+k`
`y=a (x-3)^2+2.5`

Evaluate for 'a' by substituting
`y=2` and `x=0` as mentioned above.
We get `a=-1/18`

So `y=-1/18 (x-3)^2+2.5`

Solution for (b).
Provided,
Y intercept or initial height is 2. So one point is `(0,2)` and max at x(h)=4. And range also called the total distance covered after the trajectory is 10m. While the shot is at ground height or y=0.

Hence we know,
`y=a (x-h)^2+k`
`y=a (x-4)^2+k`

Evaluate by substituting x=0 and y=2.
We get `2=16a+k`..... `color (red)((1))`
Again substitute `x=10` and `y=0` and evaluate
We get
`0=36a+k`.....`color (red)((2))`

Solving both system of equations
`color (red)((1))` and `color (red)((2))`

We get `a=-1/10` and `k=3.6`

Hence,
`y=-1/10 (x-4)^2+3.6`
Thank you.