A football is punted so that it will have a "hang time" of 4.5sec and land 45.7m away...?

Question

What is its initial velocity??

Answer 1

here's a try at it

d=s*t
45.7=s*4.5
forward speed = 10.15m/s

s-9.81t/2 will give you how fast the ball had to be to go up and down in 4.5 sec
22.05m/s
I believe this is correct, but---- I might be wrong. It's been 12 years

now we have to determine how fast and the angle of trajectory
to go up at 22.05m/s and go forward at 10.15m/s

therefore, the ball was thown at 32.2m/s at a 61.63 degree angle

proof

(61.63/90)*32.2 = 22.05m/s up
the reverse
(28.37/90)*32.2=10.15m/s forward

Answer 2

Well if you assume that its initial velocity was at an angle of 45 degrees above the horizontal its not hard.

That means its horizontal and vertical velocities are the same. And ignoring air friction it's horizontal velocity doesn't change.

So the horizontal velocity is 45.6 m/ 4.5 sec = m/s

Since the vertical velocity is the same (45 degree angle)

It's inital velocity is Horizontal velocity = initial x cos (45)

So inital velocity = Horizontal velocity / cos(45)

Answer 3

first of all you need to understand that hang time has nothing to do with how far the punt goes. hang time is only important because it allows the defense to arrive at the same time the ball lands in the hands of the receiving team.

Answer 4

You don't say what the angle is... if you do 47.5/4.5 you'll; get the veloc. on the x axis, don't know thew angule so can't determine the vectorial initial veleocity you have only vcos of the angule

Answer 5

velocity is = to distance over time so it would be 45.7m/4.5s which equals 10m/s (using correct sf)