Physics Projectile Motion Question.Showing work is appreciated.?

Question

Guy throws football at 24m/s^2 at a 45 degree angle.It takes 3seconds to reach the top of its path.The ball is caught at same height that it was thrown. How long is it in the air?

Answer 1

The time and distance are related by the equation
S = [1/2] gt^2

The vertical distance S = [1/2] g [9]

While falling down it travels the same distance S.

Time to fall down t^2 = 2S / g = 2 [1/2] g [9] /g = 9.
t = 3 s.

Total time = 3 + 3 = 6 s

Answer 2

[NOTE: based on the data in the problem, I am assuming the problem is from a very basic physics class. For example, the numbers in the problem do not suggest the correct value for gravity near Earth, etc. For this reason I am solving the problem in a very general way with NO TRIG.]

If there is no air resistance, the flight of an object makes a perfectly symmetric parabola.

So, if it took 3 seconds for gravity to slow the ball down enough to make it stop going up, it will take 3 more seconds for gravity to return it to the height it was thrown.

Here are the numbers behind the justification:

Let g=acceleration downward due to gravity.
Let v_u = upward velocity [not total velocity].

It takes 3 seconds for the ball to stop going up, so v_u = 3g.

During that time the AVERAGE velocity upward is v_u/2 , so the height it reaches is equal to the TIME [3 seconds] times the average velocity [v_u/2], so

Height it reaches = 3*v_u/2 = 3*3g/2 = 9g/2.

Let T= time it takes to drop.

Its downward velocity when it is at its peak is 0 [it is not going up or down at that instant].

It's final velocity will be Tg [time it is dropping times the acceleration due to gravity.]

So its AVERAGE velocity while it is dropping is Tg/2

It is dropping for T seconds, so the distance it drops is time times average downward velocity: T*(Tg/2) = T^2g/2

Since the distance it goes up equals the distance it goes down, we can set this equal to the height it went up:

9g/2 = T^2g/2 ---> T=3

It takes it 3 seconds to drop. So it is in the air for 6 seconds.

Answer 3

Vertical speed Vv = 24 m/s * sin(45 deg)
t(up) = t(down) = Vv/g
t(total) = t(up) + t(down)
EDIT: David and Pearl are correct. Either the problem is "overdefined" and has conflicting data (g=5.7 m/s^2), or the throw was not made at Earth's surface, or "24 m/s^2" is not a typo and is the average accel during the throw (and is irrelevant). That'll teach me to read the whole question!