A quarterback throws a pass at an angle of 40 degrees above the horizontal with an initial speed of 20 m/s. The ball is caught by the receiver 2.55 seconds later. Determine horizontal velocity, vertical initial velocity, the distance ball was thrown, and maximum height of the ball?

well, the vertical position at time t is

y(t) = 20 sin40° t - 4.9t^2
= 12.86t - 4.9t^2

I think that should get you started ok.

To solve this problem, we can break down the motion of the ball into horizontal and vertical components. Let's first analyze the horizontal motion.

1. Horizontal Velocity (Vx):
The horizontal velocity remains constant throughout the motion because there is no acceleration acting horizontally. Therefore, the horizontal velocity is the same as the initial velocity. Given that the initial speed is 20 m/s and the angle is 40 degrees above the horizontal, we can find the horizontal velocity using trigonometry:

Vx = initial speed × cos(angle)
Vx = 20 m/s × cos(40 degrees)
Vx ≈ 15.3 m/s

Therefore, the horizontal velocity is approximately 15.3 m/s.

Next, let's calculate the vertical motion components.

2. Vertical Initial Velocity (Vy):
The vertical velocity changes due to the acceleration due to gravity. In the absence of air resistance, the vertical velocity at any given time can be determined using the following formula:

Vy = initial speed × sin(angle)
Vy = 20 m/s × sin(40 degrees)
Vy ≈ 12.9 m/s

Therefore, the vertical initial velocity is approximately 12.9 m/s.

Now, let's determine the distance the ball was thrown.

3. Distance (d):
To calculate the distance, we need to find the time it takes for the ball to reach the receiver. Given that the receiver catches the ball 2.55 seconds later, we can use the horizontal velocity to find the distance:

d = horizontal velocity × time
d = 15.3 m/s × 2.55 s
d ≈ 39.0 m

Therefore, the distance the ball was thrown is approximately 39.0 meters.

Lastly, let's find the maximum height of the ball.

4. Maximum Height (H):
To determine the maximum height, we need to find the time it takes for the ball to reach its highest point. The vertical motion can be analyzed using the kinematic equation:

Vy = initial velocity + acceleration × time

Since the ball reaches its highest point when the vertical velocity becomes zero, we can rearrange the equation:

0 = Vy - g × t_max_height

Solving for t_max_height:

t_max_height = Vy / g
t_max_height ≈ 12.9 m/s / 9.8 m/s^2
t_max_height ≈ 1.32 s

Now, we can calculate the maximum height using the vertical velocity and time:

H = Vy × t_max_height - (1/2) × g × t_max_height^2
H ≈ 12.9 m/s × 1.32 s - (1/2) × 9.8 m/s^2 × (1.32 s)^2
H ≈ 8.4 m

Therefore, the maximum height of the ball is approximately 8.4 meters.