A basketball shot is taken from a horizontal distance of 5m from the hoop. The height of the ball can be modelled by the relation h = - 7.3t^2 + 8.25t + 2.1, where h is the height in meters, and t is the time in seconds, since the ball was released

This is the concept of application of quadratic expressions. Given that the height of the ball is modeled by the equation;

h=-7.3t^2+8.25t+2.1+5

The time taken for the ball to hit the ground will be given as falls;

-7.3t^2+8.25t+7.1=0

to solve for t we use the quadratic formula;

t=[-b+/-sqrt (b^2-4ac) ] / (2a)

a=-7.3, b=8.25, c=2.1

t=[-8.25+/-sqrt[8.25^2+4*7.3*7.1] / (-2*7.3)

t = - 0.572

or

t=1.702

since there is not negative time we take the time taken for the ball to hit the ground will be: t=1.702 sec