A football is thrown down the field by a quarterback from a height of 1 foot with an initial velocity of 40 feet per second at angle of pi/6 radians. If a receiver catches the ball downfield at a height of 1 foot above ground;

Question

a) How long was the ball in the air?
b) What was the maximum height of the ball?

Answer 1

recall that the height
y = 1 + 40*sin(pi/6) t - 8t^2

(a) solve for t when y=1
(b) as with all parabolas, the vertex is at t = -b/2a

Answer 2

the given height equation is incorrect
... the t^2 coefficient is -16, not -8

solve away...

Answer 3

correct. good catch