Question 177806
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 metres, and t is the time, in seconds, since the ball was released.
:
a) From what height was the ball released?
Released when t = 0, therefore -7.3(0^2) + 8.25(0) + 2.1 = 2.1 meters
:
b) What was the maximum height reached by the ball?
Max height occurs at the axis of symmetry, x = -b/(2a; in this equation we have:
t = {{{(-8.25)/(2*-7.3)}}}
t = .565 sec
Max height = -7.3(.565^2) + 8.25(.565) + 2.1,
h = -2.33 + 4.66 + 2.1
h = 4.43 meters max height
:
c) If the ball reached the hoop in 1 s, what was the height of the hoop?
Substitute 1 for t, find h
h =-7.3(1^2) + 8.25(1) + 2.1
h =-7.3 + 8.25 + 2.1
h = 3.05 meters