Abasketball 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

Aperecent is the answer