A quarterback throws a football toward a receiver with an initial speed of 18 m/s, at an angle of 29 degrees above the horizontal. At that instant, the receiver is 19 m from the quarterback. With what constant speed should the receiver run in order to catch the football at the level at which it was thrown? The acceleration of gravity is 9.8 m/s2 . Answer in units of m/s.

Vo = 18m/s @ 29deg.

Xo = hor. = 18cos29 = 15.74m/s.
Yo = ver. = 18sin29 = 8.73m/s.

Dh = Vo^2*sin(2A) / g,
Dh = (18)^2*sin58 / 9.8 = 28m = Hor. dist. of ball.

d = Vt,
V = d/t = (28-19) / 1.782 = 5.1m/s. =
Velocity of receiver.

The following steps should follow the Dh calculation:

Tr = (Yf - Yo) / g,
Tr = (0 - 8.73) / -9.8 = 0.891s = Rise
time = Time to reach max height.

Tf = Tr = 0.891s. = Fall time.

t = Tr + Tf = 0.891 + 0.891 = 1.782s =
Time in flight.

To solve this problem, we can split the initial velocity of the football into its horizontal and vertical components.

The horizontal component of the velocity (Vx) remains constant throughout the motion because there is no horizontal acceleration. So, Vx = 18 m/s.

The vertical component of the velocity (Vy) changes due to the effect of gravity. To find Vy, we use the equation:

Vy = V * sin(theta)

Where V is the initial speed and theta is the angle above the horizontal. Substituting the values given:

Vy = 18 m/s * sin(29 degrees)
Vy ≈ 9.32 m/s

Now, let's find the time it takes for the football to reach the receiver. We can use the equation:

y = Vyi * t + 1/2 * a * t^2

Since the initial vertical velocity is Vy = 9.32 m/s and the vertical acceleration is -9.8 m/s^2 (due to gravity), we can rearrange the equation:

19 m = 9.32 m/s * t + 1/2 * (-9.8 m/s^2) * t^2

Simplifying the equation and moving all terms to one side:

1/2 * (-9.8 m/s^2) * t^2 + 9.32 m/s * t - 19 m = 0

To solve this quadratic equation, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Using the formula, we find two possible values for t: t ≈ 1.051 s and t ≈ 2.19 s. Since we are looking for the time of flight of the ball until it reaches the receiver, we use t = 2.19 s.

Now, to find the required speed of the receiver, we can use the horizontal distance formula:

x = Vx * t

Substituting the values, we have:

19 m = 18 m/s * 2.19 s

Simplifying, we find:

19 m = 39.42 m

To catch the ball at the same level it was thrown, the receiver should run with a constant speed of 39.42 m/s.

To solve this problem, we need to find the horizontal component of the velocity of the football and then calculate the time it takes for the football to reach the receiver at a distance of 19m.

1. Find the horizontal component of the initial velocity:
The horizontal component can be calculated using the formula: vx = v * cos(θ)
where vx is the horizontal component of velocity, v is the initial speed (18 m/s), and θ is the angle of 29 degrees.
Plug in the values: vx = 18 * cos(29) = 15.89 m/s

2. Calculate the time it takes for the football to reach the receiver:
The time can be calculated using the formula: t = d / vx
where t is the time, d is the distance to the receiver (19m), and vx is the horizontal component of velocity (15.89 m/s).
Plug in the values: t = 19 / 15.89 = 1.195 seconds

3. Now, we know that the receiver needs to run at a constant speed to keep up with the football in order to catch it at the level at which it was thrown. This means that the receiver needs to cover the same horizontal distance as the football in the same time. Therefore, we can calculate the speed the receiver needs to run using the formula: s = d / t
where s is the required speed, d is the distance to the receiver (19m), and t is the time (1.195 seconds).
Plug in the values: s = 19 / 1.195 = 15.9 m/s

So, the receiver needs to run at a constant speed of 15.9 m/s in order to catch the football at the level at which it was thrown.