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?

 

[rwisz]

I would first clarify the variables and assumptions in the problem. It is important to understand what is meant by "thrown deep into the end zone" and whether the ball's initial velocity is known. I would also confirm with the person giving the problem that air friction can indeed be neglected.

Assuming that the initial velocity is known and air friction can be neglected, the kinematic equation you used is correct. However, the displacement you calculated is not the height of the ball. It is the total distance traveled by the ball during its flight, including both the horizontal and vertical components.

To find the maximum height, we need to consider the vertical component of the ball's motion. Since the ball is thrown and caught in the same location, we know that the vertical velocity at the highest point is 0 m/s. We can use this information to find the maximum height using the following equation:

v_f^2 = v_i^2 + 2ad

Where v_f is the final velocity (0 m/s), v_i is the initial vertical velocity (unknown), a is the acceleration due to gravity (-9.81 m/s^2), and d is the displacement (43.3 m). Solving for v_i, we get:

v_i = √(2ad)

Plugging in the values, we get:

v_i = √(2(-9.81)(43.3)) = 29.4 m/s

Now, we can use this initial vertical velocity in the kinematic equation to find the maximum height:

v_f = v_i + at

0 = 29.4 + (-9.81)t

t = 3 seconds

Finally, we can use this time in the kinematic equation for displacement to find the maximum height:

d = v_i(t) + 1/2at^2

d = 29.4(3) + 1/2(-9.81)(3)^2

d = 44.1 m

Therefore, the maximum height reached by the football is approximately 44.1 meters.