A goalie slips and falls to the ground and the opponent's forward, 20 m out in front of the net boots the ball toward the net at a speed of 15.9m/s and an angle of 45 degrees above the horizontal. Does the ball go into the net?

The ball (neglecting spin and drag effects) travels

(V^2/g) = 25.7 m before hitting the ground. At the net, its height must be considered to make sure it does not go over the net. It can certainly get there.

Travel time time to the net
t = 20 m/(15.9 cos 45)= 1.78 s

ball height after reaching net
= 11.24 t - (g/2)t^2 = 4.5 m

That is higher than the soccer goal height of 8 feet = 2.6 meters

So it sails over the goal.

Didn't they tell you the soccer goal height?

The crossbar of the goal is 2.5 m above the ground. Thanks so much!

To determine whether the ball goes into the net, we need to analyze its trajectory and determine if it crosses the goal line.

To solve this problem, we can break down the initial velocity of the ball into its horizontal and vertical components. The horizontal component remains constant, while the vertical component is affected by gravity.

Let's calculate the time it takes for the ball to reach the net. We can ignore air resistance for this analysis.

The vertical component of the initial velocity can be found by multiplying the initial velocity (15.9 m/s) by the sine of the launch angle (45 degrees):

Vertical component = 15.9 m/s * sin(45 degrees)

Using the trigonometric identity sin(45 degrees) = √2/2:

Vertical component = 15.9 m/s * √2/2
≈ 15.9 m/s * 0.7071
≈ 11.27 m/s

We can use this value to calculate the time it takes for the ball to reach its peak height (maximum height). At this point, the vertical velocity becomes zero.

Using the equation vf = vi + at, where vf is the final vertical velocity, vi is the initial vertical velocity, a is the acceleration due to gravity (-9.8 m/s^2), and t is the time:

0 = 11.27 m/s + (-9.8 m/s^2) * t_peak

Solving for t_peak:

-11.27 m/s = -9.8 m/s^2 * t_peak
t_peak = -11.27 m/s / (-9.8 m/s^2)
t_peak ≈ 1.15 s

Since the time to reach the peak is the same as the time to fall back to the original height, the total time of flight, t_total, is approximately twice the time to reach the peak:

t_total = 2 * t_peak
t_total ≈ 2 * 1.15 s
t_total ≈ 2.3 s

Now, let's calculate the horizontal distance the ball travels during this time. We use the horizontal component of the initial velocity, which remains constant:

Horizontal distance = horizontal component * time of flight
= 15.9 m/s * cos(45 degrees) * 2.3 s

Using the trigonometric identity cos(45 degrees) = √2/2:

Horizontal distance ≈ 15.9 m/s * √2/2 * 2.3 s
≈ 15.9 m/s * 0.7071 * 2.3 s
≈ 23.19 m

The ball travels approximately 23.19 meters horizontally during its flight.

Now, we know that the forward is 20 meters out in front of the net. If the horizontal distance traveled by the ball is greater than or equal to 20 meters, it means the ball goes into the net.

Since 23.19 meters is greater than 20 meters, the ball goes into the net.