A 80kg football player leaps straight up into the air (with no horizontal velocity) to catch a pass. He catches the .43kg ball at the peak of his jump, when he is 17.2 inches off the ground. He hits the ground 1.61 inches away from where he leapt. If the ball was moving horizontally when it was caught how fast was the ball traveling?

How long does it take to fall 17.2 inches.

(I assume you are working with an ancient text that uses feet and inches so in your world g = 32 ft/s^2)
17.2 inches /12 in/ft = 1.43 ft

d = 1.43 = (1/2) g t^2
1.43 = 16 t^2
t = .299 seconds to fall

So horizontal speed u took us 1.61 inches in .299 seconds
1.61/12 = u (.299)
so u = .449 ft/second

Call ball speed s
total x momentum of system before catch = momentum of ball = .43 s
so
.43 s = u(80.43) = .449 (80.43)
so
s = 84 ft/s

To determine the speed at which the ball was traveling horizontally, we can use the conservation of momentum. The momentum before the catch should be equal to the momentum after the catch.

Let's break down the problem step by step:

Step 1: Determine the initial vertical velocity of the player
The player leaps straight up into the air with no horizontal velocity. The initial vertical velocity is 0 m/s.

Step 2: Determine the final vertical velocity of the player
To find the final vertical velocity, we can use the kinematic equation:
vf^2 = vi^2 + 2ad
where vf is the final vertical velocity, vi is the initial vertical velocity, a is the acceleration, and d is the distance.
In this case, the player reaches a maximum height of 17.2 inches (converted to meters) and then falls back down to the ground, so the distance traveled is twice the maximum height (34.4 inches). The acceleration due to gravity is 9.8 m/s^2 (downward).
Plugging in the values:
0^2 = vi^2 + 2 * (-9.8) * 0.44
0 = vi^2 - 8.624
vi^2 = 8.624
vi = √8.624
vi ≈ 2.94 m/s (upward)

Step 3: Determine the final vertical velocity of the ball
Since the ball was caught at the peak of the player's jump, both the player and the ball have the same final vertical velocity. Therefore, the vertical velocity of the ball is also 2.94 m/s (upward).

Step 4: Determine the time it takes for the player to reach the peak
We can use the equation:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
Plugging in the values:
2.94 = 0 + (-9.8) * t
2.94 = -9.8t
t = 2.94 / 9.8
t ≈ 0.3 seconds

Step 5: Determine the horizontal distance traveled by the ball
To find the horizontal distance traveled by the ball, we can use the equation:
d = v * t
where d is the distance, v is the horizontal velocity, and t is the time.
In this case, d is given as 1.61 inches (converted to meters) and t is 0.3 seconds.
Plugging in the values:
1.61 = v * 0.3
v = 1.61 / 0.3
v ≈ 5.37 m/s (horizontal velocity of the ball)

Therefore, the ball was traveling horizontally at approximately 5.37 m/s.