Football thrown, when does it reach max.height?

[aeromat]

Homework Statement

Football is thrown deep into the end zone for a touchdown. If the ball was in the air for 2.1s and air friction is neglected, to what vertical height must it have risen?

Homework Equations

Not really sure; kinematics equations?

The Attempt at a Solution

I wasn't sure how to get height, so I tried getting displacement to see if it would lead me anywhere:

d = v(t) +1/2(g)(t)^2
d = 0 +1/2(g)(t)^2
d = +1/2(9.81)(2.1)^2
d = 43.3m

From the displacement, I don't know how I can possibly retrieve the height. Would anyone mind helping me out here?

 

[SammyS]

Assume it's caught at the same height it had when released.

During half it's flight, it's rising. During half it's dropping.

What is the vertical at the highest point in the ball's flight?

How much time elapses from the highest point in the ball's flight to the time the ball is caught?

How far will the ball fall in that amount of time?

Remember: You can treat horizontal & vertical motion independently if you ignore air resistance.

 

[gneill]

The vertical component of motion is independent of the horizontal component. If the ball was in the air for 2.1s, then its motion in the vertical direction was such that it rose high enough to be in the air for that long.

If you were to throw an object vertically upward and then catch it on its return, and it stayed in the air for 2.1s, how high would you have had to have thrown it?

 

[ileacus]

Your problem statement is either missing some information or requires some assumptions. If you assume that the receiver catches the ball at exactly the same height that the quarterback released it, and if you take that height to be 0, then this is solvable.

The displacement in this case is height, not the horizontal displacement. To begin, you've done the math a bit wrong. 0.5 x 9.81 x (2.1)^2 = 21.63m. You forgot to multiply by 1/2. (Consider what it would look like for a quarterback to throw a ball 43.4m (142 feet) in the air!) But, you also need to consider that the ball travels upwards for some time to its maximum height and then travels downwards to the receiver.

With the two assumptions above and neglecting air resistance, you can infer that half of the 2.1s of air time were spent going up, and the other half going down. At what height (above the receiver's hands) would the ball spend 1.05s in the air?