A quarterback makes a pass, throwing the ball at 21 m/s and an angle of 60° above the horizontal.

A) What is the maximum height of the ball in flight?
B) How long does it take to complete the pass?
C) How many yards (yd) does he manage to throw the ball? (1 yard = 0.9144 meters)
D) The receiver has a 22 yard lead from the quarter back by the time he throws the ball. How fast must he run in order to catch the ball when it lands?

To solve these problems, we can use the equations of projectile motion. The key variables are the initial velocity (v0), angle (θ), time of flight (t), maximum height (h), and horizontal distance (d).

A) To find the maximum height of the ball in flight, we can use the equation:

h = (v0^2 * sin^2(θ)) / (2 * g)

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Let's calculate it:

h = (21^2 * sin^2(60°)) / (2 * 9.8)
= (441 * (sqrt(3)/2)^2) / 19.6
= (441 * 3/4) / 19.6
= 1102.25 / 19.6
≈ 56.22 meters

Therefore, the maximum height of the ball is approximately 56.22 meters.

B) To find the time of flight, we can use the equation:

t = (2 * v0 * sin(θ)) / g

Let's calculate it:

t = (2 * 21 * sin(60°)) / 9.8
= (42 * (sqrt(3)/2)) / 9.8
= 21 * sqrt(3) / 9.8
≈ 2.28 seconds

Therefore, it takes approximately 2.28 seconds for the ball to complete its flight.

C) To find the horizontal distance, we can use the equation:

d = v0 * cos(θ) * t

Let's calculate it:

d = 21 * cos(60°) * 2.28
= 21 * 0.5 * 2.28
= 24.57 meters

Now, we need to convert the distance from meters to yards:

d_yard = d / 0.9144
= 24.57 / 0.9144
≈ 26.85 yards

Therefore, the quarterback manages to throw the ball approximately 26.85 yards.

D) To find the necessary running speed for the receiver to catch the ball when it lands, we need to calculate the time it takes for the ball to reach the receiver after it lands.

The time is given by the equation:

t_receiver = t + t_lead

where t_lead is the time it takes for the receiver to cover the initial 22 yards lead.

Let's calculate t_lead:

t_lead = d_lead / v_receiver

where d_lead is the 22 yard lead and v_receiver is the speed of the receiver.

Converting the lead distance to meters:

d_lead = 22 * 0.9144
≈ 20.12 meters

Now, let's calculate t_lead:

t_lead = 20.12 / v_receiver

Since the receiver starts running when the ball is thrown, they both travel the same time t_lead when the ball lands. Therefore:

t + t_lead = t_receiver

Substituting the known values:

t_receiver = t + (20.12 / v_receiver)

Since we know t (2.28 seconds) from Part B, we can solve for v_receiver:

v_receiver = 20.12 / (t_receiver - 2.28)

Note that we need to choose a value for t_receiver, which is the total time it takes for the receiver to catch the ball after it is thrown. Let's assume t_receiver = 3 seconds:

v_receiver = 20.12 / (3 - 2.28)
= 20.12 / 0.72
≈ 27.89 m/s

Therefore, the receiver needs to run approximately 27.89 m/s to catch the ball when it lands.

To solve these questions, we can use the kinematic equations of motion and apply them to the projectile motion of the thrown ball.

Let's break down the questions one by one:

A) What is the maximum height of the ball in flight?

To find the maximum height, we can use the equation for vertical motion. The initial vertical velocity (Vy) is given by V * sinθ, where V is the magnitude of the initial velocity and θ is the angle above the horizontal.

Vy = V * sinθ = 21 m/s * sin(60°) ≈ 18.14 m/s

The time taken to reach the maximum height can be determined using the equation Vy = Voy - gt, where Voy is the initial vertical velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.

0 = 18.14 - 9.8t

Solving for t:

t ≈ 1.85 seconds

Substituting this value of t into the equation for vertical displacement, we can find the maximum height:

y = Voy * t - (1/2)gt^2

y = 18.14 * 1.85 - (1/2) * 9.8 * (1.85)^2 ≈ 18.54 meters

So, the maximum height of the ball in flight is approximately 18.54 meters.

B) How long does it take to complete the pass?

To find the total time, we need to consider the time taken for the ball to reach its maximum height, and then the time taken for it to return to the ground.

The time taken to reach the maximum height is the same as the time calculated in part A, which is approximately 1.85 seconds.

To calculate the time taken for the ball to return to the ground, we need to consider the vertical motion of the ball. The equation for vertical position is y = Voy * t - (1/2)gt^2; since the ball returns to the same vertical position as it started, y is zero and we can solve for t:

0 = 18.14t - (1/2) * 9.8t^2

Solving this quadratic equation, we find two valid solutions: t = 0 and t ≈ 3.74 seconds.

However, t = 0 indicates the initial timing, so we can discard this solution.

Therefore, the total time taken for the pass is approximately 1.85 seconds + 3.74 seconds ≈ 5.59 seconds.

C) How many yards (yd) does he manage to throw the ball?

We know that 1 yard is equal to 0.9144 meters.

To find the horizontal displacement, we can use the equation for horizontal motion. The initial horizontal velocity (Vx) is given by V * cosθ, where V is the magnitude of the initial velocity and θ is the angle above the horizontal.

Vx = V * cosθ = 21 m/s * cos(60°) ≈ 10.5 m/s

The horizontal displacement (x) can be calculated using the formula x = Vx * t:

x = 10.5 m/s * 5.59 s ≈ 58.60 meters

Converting the horizontal displacement to yards:

58.60 meters * (1 yard / 0.9144 meters) ≈ 64.10 yards

Therefore, he manages to throw the ball approximately 64.10 yards.

D) The receiver has a 22 yard lead from the quarterback by the time he throws the ball. How fast must he run in order to catch the ball when it lands?

To determine the speed at which the receiver must run, we need to calculate the horizontal component of the receiver's velocity.

The horizontal displacement required by the receiver in the time taken for the pass is 58.60 meters.

Let's assume the time taken by the ball to reach the receiver is the same as the time calculated in part B, which is approximately 5.59 seconds.

Since the receiver has a 22-yard head start, the total horizontal distance covered by the receiver is 58.60 meters + 22 yards ≈ 80.19 meters.

To find the required velocity, we can use the formula Vx = x / t:

Vx = 80.19 meters / 5.59 seconds ≈ 14.35 m/s

Therefore, the receiver must run at a speed of approximately 14.35 m/s to catch the ball when it lands.