the height is
h = 1 + (21.5 sin35°)t - 4.9t^2
h=0 when t=2.6
the horizontal speed is a constant 21.5 cos35° m/s
so how far did it go in 2.6 seconds
the vertical speed is 21.5 sin35° - 9.8t
so now you can figure the velocity on impact
A football is punted and leaves the punter’s foot at a height of 1.00 m above the ground. The football’s initial velocity is 21.5 m/s 35o above the horizontal.
(a) How far did the football travel horizontally before hitting the ground?
(b) What is the football’s velocity upon contact with the ground? Specify magnitude and direction
1 answer