Question
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;
a) How long was the ball in the air?
b) What was the maximum height of the ball?
a) How long was the ball in the air?
b) What was the maximum height of the ball?
Answers
Steve
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
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
Scott
the given height equation is incorrect
... the t^2 coefficient is -16, not -8
solve away...
... the t^2 coefficient is -16, not -8
solve away...
Steve
correct. good catch