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. ?

1 Answer
AJ
Apr 18, 2018

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.

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